David Naccache
email: david.naccache@ens.fr

Ofer Yifrach-Stav

The Balkans Continued Fraction

In a previous escapade we gave a collection of continued fractions involving Catalan's constant. This paper provides more general formulae governing those continued fractions. Having distinguished different cases associated to regions in the plan, we nickname those continued fractions "The Balkans" as they divide into areas which are related but still different in nature. Because we do not provide formal proofs of those machine-constructed formulae we do not claim them to be theorems. Still, each and every proposed formula was extensively tested numerically.

1 Each snippet ends by a ClearAll["Global' * "]; command whose purpose is to make Mathematica "forget" all passed history.

Introduction

In a previous escapade [12] we gave a collection of continued fractions involving Catalan's constant. This paper provides more general formulae governing those continued fractions. Having distinguished different cases associated to regions in the plan, we nickname those continued fractions "The Balkans" as they divide into areas which are related but still different in nature.

Because we do not provide formal proofs of those machine-constructed formulae we do not claim them to be theorems. Still, each and every proposed formula was extensively tested numerically.

All the programs included in this article are self-contained, i.e. any code snippet can be run independently of the others to fully illustrate the encoded formula. This renders the code longer but has the great advantage of allowing the reader to run and modify each snippet directly by just cutting and pasting it into Mathematica without requiring any other module 1 . The code was compacted for the sake of concision but loading it into Mathematica's editor re-indents it automatically.

The code in this paper is necessary. Because we do not provide proofs explaining the discovered structures, any slight L A T E X misprint would be impossible to fix. Hence readers can consider the code as an unambiguously tested version of the proposed formulae.

Notations

We denote by n!! the semifactorial of, i.e. the product of all the integers from 1 up to n having the same parity as n:

n!! = n 2 -1 k=0 (n -2k) = n(n -2)(n -4) • • •
Because in all the following we will only apply semifactorials to odd numbers, this can be simplified as:

n!! = n+1 2 k=1 (2k -1) = n(n -2)(n -4) • • • 3 • 1
We denote by Catalan's constant by G = 0.91596559 . . . and let C n be the n-th Catalan number:

C n = 1 n + 1 2n n = (2n)! (n + 1)! n! = n k=2 n + k k for n ≥ 0
The first Catalan numbers are:

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, . . .

The target

We define for odd j and κ, c ∈ N the following quantity nicknamed "The Balkans continued fraction":

Q j,κ,c = j(2 -j + 2κ) + ∞ K n=1 -2n(c + n)(j + n -1)(1 -j + 2κ + n) j(2 -j + 2κ) + (3 + 4κ)n + 3n 2
The question asked is that of finding a general process allowing to compute Q j,κ,c without resorting to numerical simulations or integer relation algorithms. The reason for this is that while integer relation algorithms allow us to "magically" discover relations, they do not provide general information about the underlying structure of the constants found. Why? Because we're in it for the thrill of discovery, not just the magic of shortcuts.

The contribution

The main contribution of this paper is a collection of formulae computing Q j,κ,c without requiring any numerical simulation for positive j, κ, c and odd j.

Why this formula in particular?

The Ramanjuan Machine Project [14,5], lists a few continued fractions involving G detected in 2020. We do not know why the project did not resort to (rather basic) integer relation algorithms to discover more relations. We hence decided to play detective and unleashed LLL that found a few hundreds of continued fractions involving G in a few intensive calculation days. All continued fractions were of the form:

+ ∞ K n=1 -2n(n + τ)(n + η)(n + µ) + δn + 3n 2 = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z

We performed two natural tests on the coefficient vectors (δ, , τ, η, µ): a PCA to determine if the coefficients can be expressed as linear combinations of less than 5 variables and a Hough transform to detect affine relations in the dataset.

PCA revealed that, when projected on (δ, τ, η, µ), nearly all data was governed by three linear dimensions 2 . We hence understood that we were facing a linear behavior in a large region (Balkans) plus some sporadic cases (see Remark 12). This was also confirmed by the Hough transform that detected several parallel plans in the 3D-space. We thus decided to focus our efforts on the main plans in the 3D-space and understand them.

How formulae were reverse-engineered

Now comes the real adventure. The process that allowed us to reverse-engineer the formulae given in this paper is interesting by its own right. A quick look at many examples of the three quantities a 0 , a 1 , a 2 forming the fractions:

Q j,κ,c = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z showed that the a i s are products of small prime factors and a few large prime factors. This suggested that a i s were initially 3 of some form: (j, κ, c) where the φ i are functions such as (an+b)!, 2 an+b , (an+b)!!, C an+b , Pochhammer symbols of linear combinations of the parameters j, κ, c etc. and a few unknown "mixing" functions causing the appearance of the large prime factors, e.g. polynomials or recurrence relations.

a i = expression(j, κ, c) = u-1 i=0 φ i (j, κ, c) v-1 i=0 φ i
Fortunately, integer relation algorithms allow us to collect many instances of such forms for diverse j, κ, c values. Hence the problem at hand consists in identifying which φ i s are compatible with the cancellations due to the division. 2 To which we gave the names j, κ, c. 3 i.e., before simplification intervenes.

If a given φ is present in the expression then it is reasonably assumed that when tried for many j, κ, c the new expression: expression 1 (j, κ, c) = expression(j, κ, c) φ (j, κ, c) or expression 1 (j, κ, c) = expression(j, κ, c) • φ (j, κ, c) will feature less small factors and hence stand-out as an outlier. The process can hence be repeated with proper backtracking until all the combinatorial φ i s were peeled-off. Then it remains to detect what the remaining "mixing" functions are which is done by monitoring the average growth rate of those surviving constants to emit hypotheses on the type of recurrence relations (or polynomials) at hand or resorting to a variety of integer sequence recognition tools to identify the hidden culprits.

-We started our exploration with the simplest case of Bosnia & Herzegovina where a 2 = 0. We chose Bosnia & Herzegovina as a launchpad because it is the simplest finite Balkan continued fraction. We (rightfully) hoped that analyzing it will give us insight about the Balkans' structure before hell breaks loose with infinite continued fractions spitting Gs into the convergence values. -Having reverse-engineered Bosnia & Herzegovina we moved to Croatia for which a 2 = 0 as well. Because Croatia is a more complex version of Bosnia & Herzegovina, the insight gained in Bosnia & Herzegovina proved very helpful to derive further characteristics of Croatia . We thus stress that the distinction between Bosnia & Herzegovina and Croatia is paedagogic rather than scientific. -The insight gained in Croatia guided our software to the formula for which is simpler than Kosovo and Serbia given that Montenegro corresponds to j = 1. -Having inferred Montenegro we moved on to Kosovo whose symmetry 4 with Serbia was quickly noted.

This work demonstrates the interest of statistical classifiers such as Maximum Likelihood Estimation (MLE) and Support Vector Machines (SVM) in mathematical exploration.

As will be shown, this "gradient descent" method proved itself very well, although it required a few thousands of computation hours on a very powerful cluster.

We estimate that 80% of the discovery effort was done by the machine. The remaining 20% being human "piloting" that, we are convinced, is already at the reach of today's most powerful LLMs, such as Gemini.

It appears much better to read the coming sections first to understand what prey we are stalking and then refer to Section 12 describing the hunting process.

4 Kosovo, Serbia, Croatia, Montenegro, Bosnia & Herzegovina .

As we will see, we distinguish five cases that we call Kosovo, Serbia, Croatia, Montenegro and Bosnia & Herzegovina after the regions of interest in the (j, κ)space shown in Figure 1. In each region c runs over the integers.

Area Domain Q j,κ,c is in Croatia j ≥ 2κ + 5 Q Bosnia & Herzegovina j = 2κ + 3 Q Serbia 2κ + 1 ≥ j ≥ κ + 3 R Kosovo κ + 2 ≥ j ≥ 3 R Montenegro j = 1 R Table 1. Areas. We list fractions involving G as being in R although it is currently un- known if G ∈ Q (this is a major open question).
A first restriction of our study will be to focus on odd j that produce Q j,κ,c values involving G. Note that:

Q j,c,κ =              a 0 a 1 + a 2 G if j is odd a 0 a 1 + a 2 log 2 if j is even
Note as well that j, κ, c can be negative. In the examples given in the appendix we adopt the notation:

a 0 a 1 + a 2 • constant = T (0) + ∞ K n=1 P (n) T (n)
This paper does not treat the even j case (a follow-up paper dealing with those cases is underway). Negative j, κ, c are of little interest as the Balkan's numerator is defined as -2n(c + n)(j + n -1)(1j + 2κ + n) hence, if c < 0 the term (c + n) will hit zero for n = -c and subsequently render the summation finite. The same happens for negative j with the term (j + n -1) and for negative κ with the term (1j + 2κ + n).

The Montenegro Conjecture

We start with the first j, κ space called Montenegro. Montenegro corresponds to the case j = 1. In other words:

∆ κ,c (α, β) =            α + βc if c < 2 -2c(2c -1)(2(c -κ) -1) 2 ∆ κ,c-2 (α, β) if c ≥ 2 + (8c 2 + (2 -8κ)c -2κ + 1)∆ κ,c-1 (α, β) Γ κ,c (α, β) = (2c -1)!! 2 G + ∆ κ,c-1 (α, β) • κ-1 i=0 (2(c -i) -1) δ κ = 4 κ-1 (2κ -1)C κ-1 and ρ κ = δ κ (-1) κ (1 -2κ) (2κ)!(2κ -3)!! α κ = ρ κ ∆ 1,κ-1 (1, -2) and β κ = -ρ κ (2κ -3) 2 ∆ 2,κ-1 (1, 12) -α κ Then ∀κ, c ∈ N 2 , Q 1,κ,c = δ κ (2c)! Γ κ,c (α κ , β κ ) Q 1,κ,c
is hence an explicitly computable fraction of the form:

Q 1,κ,c = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z

The code snippet testing this formula over the square 1 ≤ κ, c ≤ 14 is entitled "1. Montenegro".

α Γ Q ∆ ρ β δ Fig. 3. Functional dependency between the functions computing Q 1,κ,c
In summery in Montenegro we do not need to resort to any integer relation algorithms to compute Q 1,κ,c for all κ, c values.

Remark 1. Note that, as described here, the complexity of ∆ κ,c is exponential in c, however, using classical Fibonacci memoization, this complexity can be reduced to O(c log c) thereby resulting in a very efficient algorithm for computing Q 1,κ,c .

The Bosnia & Herzegovina Conjecture

Bosnia & Herzegovina corresponds to the line 2κ = j -3, that is:

Q j, j-3 2 ,c = -j + ∞ K n=1 -2n(n -2)(n + c)(n + j -1) 3n 2 + (2j -3)n -j = 2 + 2c -j
We start by defining:

α j = 1 and β j = 15 -4j
And let:

∆ j,c (α j , β j ) =            α j + β j c if c < 2 -2c(2c -j)(2c + 1)(2c -j + 2)∆ j,c-2 (α j , β j ) if c ≥ 2 + (8c 2 + (14 -4j)c -3(j -2))∆ j,c-1 (α j , β j ) g(j, c) = (2c)! 2 j-1 2 i=1 (2 + 2i -j) and h(j, c) = j-5 2 i=1 (2c -2i -1)
Then:

Q j, j-3 2 ,c = g(j, c) ∆ j,c-1 (α j , β j ) • h(j, c) ∈ Q
Hence, in Bosnia & Herzegovina as well we do not need to resort to any integer relation algorithms to compute Q j, j-3

2 ,c which is an explicitly computable fraction in Q.

The corresponding code snippet is "2. Bosnia".

Note that Q j, j-3 2 ,c = 2+2c-j gives a non-recursive formula for ∆ j,c (1, 15-4j). This "long detour" for computing Q j,(j-3)/2,c which is a finite continued fraction (exactly computable by summation) is extremely useful as it unveiled the ∆ structure and the basic φ i s that repeatedly intervene in other Balkan areas.

Roadmap

We are now ready to describe the roadmap that will govern the rest of this paper.

Areas governed by one running variable

As we have just seen, Montenegro and Bosnia & Herzegovina follow similar behaviors. This pattern revolves around the "magic" values α, β which are formally known for Montenegro and Bosnia & Herzegovina. Both regions are lines parameterized by a single running variable (κ for Montenegro and j for Bosnia & Herzegovina).

The c-level master formula for all Balkans except Montenegro

Except Montenegro (whose case was settled) all other areas obey a common clevel master formula that we will provide below. Computing Q j,κ,c for any c using this c-level master formula requires knowledge of two rational parameters: α j,κ and β j,κ .

The cornerstone of the rest of this paper is thus the quest for α j,κ and β j,κ for all Balkan areas except Montenegro. α j,κ , β j,κ can always be inferred by computing numerically [START_REF] Cohen | Elementary Continued Fractions for Linear Combinations of ζ and L Values[END_REF] Q j,κ,c 1 and Q j,κ,c 2 for two values c 1 c 2 and solving a system of two equations in the unknowns α j,κ , β j,κ . When α j,κ , β j,κ are found Q j,κ,c can be computed for any other c {c 1 , c 2 }. Using this process requires resorting twice to integer relation algorithms such as LLL [11], HJLS [10], PSOS [1,7] or PSLQ [8] for each (j, κ)-pair. This works perfectly in practice but is not entirely satisfactory because the process has a "blind spot" which is the integer relation oracle. We would like to avoid such blind spots as much as possible and dispose of a fully algebraic process for computing each Q j,κ,c . The ideal situation being, of course, a direct algebraic computation of Q j,κ,c from the data j, κ, c alone.

We completely achieve the goal of computing all Q j,κ,c values in an exact algebraic way all over the Balkans. Note that we do not need to bother with Croatia where Q j,κ,c is a finite continued fraction computable exactly by summation. This simply stems from the fact that as we increase n the (1j + 2κ + n) component of Balkan's continued fraction will necessarily hit 0 in Bosnia & Herzegovina and in Croatia. The snippet "3. Northern Balkans" illustrates the process with all areas except Montenegro. The snippet validates the c-level master formula on the 3D volume 3 ≤ j ≤ 13, 1 ≤ κ ≤ 6, 1 ≤ c ≤ 7 (Figure 16) stretching over parts of Croatia, Bosnia & Herzegovina, Serbia and Kosovo. In each case the program derives the corresponding α j,κ , β j,κ and allows the computing of Q j,κ,c for any c. The formally computed results are then successfully compared to numerical ones.

As we will later see, for Serbia and Kosovo we have more powerful master formulae operating at the κ-level and the j-level.

The c-level master formula 6 is defined as follows, assuming that we are somehow given the magic constants α j,κ , β j,κ .

Let:

∆ j,κ,c (α j,κ , β j,κ ) =            α j,κ + β j,κ c if c < 2 -2c(2c -j)(2c -2κ + j -2)(2c -2κ -1)∆ j,κ,c-2 (α j,κ , β j,κ ) if c ≥ 2 + (8c 2 + (2 -8κ)c + (j -2)(2κ -j))∆ j,κ,c-1 (α j,κ , β j,κ )
6 valid for all areas except Montenegro whose case was anyhow previously settled.

f j,κ,c = C j-3 2 C κ-1 (j -2)(2κ -1)(2c -1)!! 2 j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1) g j,κ,c = (2c)!2 j+4κ-7 2 j-1 2 i=1 (2c -2i + 1)(2κ -2i + 1) h j,κ,c = j-3 2 i=0 (2c -2i -1) κ-1 i=0 (2c -2i -1)
Then:

Q j,κ,c = g(j, κ, c) ∆ j,κ,c-1 (α j,κ , β j,κ ) • h(j, κ, c) + f (j, κ, c) • G Remark 2.
For Bosnia & Herzegovina and Croatia f j,κ,c = 0. This happens automatically given the definition of f j,κ,c .

The Serbia conjecture

Montenegro, Kosovo and Serbia feature internal symmetries illustrated by the arches in Figures 1 where connected Q values are identical. We will also provide a formula connecting their α, β values over the plan. This means that any formula applicable to Kosovo will also settle the case of Serbia (given that Montenegro is settled and given that Montenegro provides the values for the upper border line of Serbia).

Because:

Q 1+2i,κ,c = Q 2κ+1-2i,κ,c valid for κ = 2, 3, . . . and 0 ≤ i ≤ k 2 -1
To compute any value in Serbia, we can just compute its symmetric correspondent in Kosovo.

The Kosovo Conjecture

Kosovo is very specific in that it has a κ-level master formula. This means that for each j, knowledge of four upper-level rational constants7 :

{α j , β j } = { α α j , β α j }, { α β j , β β j }
allows to generate α j,κ , β j,κ for all κ values along a j-line within Kosovo. Finding α j , β j for a given j requires solving a system of equations using four known α j,κ 1 ,β j,κ 1 ,α j,κ 2 ,β j,κ 2 for some κ 1 κ 2 . To determine the two α j,κ i , β j,κ i pairs [START_REF] Ferguson | Analysis of PSLQ, an Integer Relation Finding Algorithm[END_REF] we can solve two systems of equations (each in two unknowns) using any

Q j,κ 1 ,c 1 ,Q j,κ 1 ,c 2 ,Q j,κ 2 ,c 3 and Q j,κ 2 ,c 4 .
Note that there is no opposition to take c 1 = c 3 and c 2 = c 4 . This process is illustrated in Figure 5.

α j , β j any α j,κ , β j,κ any Q j,κ,c α j,κ 1 , β j,κ 1 Q j,κ 1 ,c 1 Q j,κ 1 ,c 2 α j,κ 2 , β j,κ 2 Q j,κ 2 ,c 3 Q j,κ 2 ,c 4
computed using an integer relation algorithm Fig. 5. κ-level resolution process for Kosovo.

The Kosovo κ-level master formula

The κ-level master formula for Kosovo allows to infer α j,κ , β j,κ from α j , β j for a fixed j and a variable κ.

In other words, the Kosovo κ-level master formula generates for a fixed j and for a variable κ the data α j,κ , β j,κ necessary to operate the general c-level master formula given in subsection 7.2.

Recall that the c-level master formula of subsection 7.2 generates for a fixed j, κ and a variable c a formal expression of the continued fraction Q j,κ,c , given the auxiliary input α j,κ , β j,κ .

Define:

π(j, κ) = j-3 2 i=0 (κ -i)(2κ -2i -1) 2 (n, j, κ) = (-1) κ+1 (2κ)! 2 κ!2 3κ-2 (2κ -j)(2κ -1)(n((2κ -j -2)(3 -2κ) -1) + 1) • π(j, κ) η(n, j, κ) = (2κ + 2j -9 -2n)(2κ + j -8 -2n)(-2κ + 5 -j)(2κ + j -6) φ(n, j, κ) = 8κ 2 + κ(10j -48 -8n) + 3j 2 -(28 + 4n)j + 68 + 18n ∆n,j,κ (α, β) =        α + βκ if κ < 2 η(n, j, κ) • ∆n,j,κ-2 (α, β) + φ(n, j, κ) • ∆n,j,κ-1 (α, β) if κ ≥ 2
We assume that we are given the four constants:

{α j , β j } = { α α j , β α j }, { α β j , β β j } ∈ Q 4
Then:

α j,κ = ∆0,j,κ-j+2 ( α α j , β α j ) (0, j, κ)
and

β j,κ = ∆1,j,κ-j+2 ( α β j , β β j ) (1, j, κ) -α j,κ
The process is illustrated by the code snippet "4. Kosovo".

The computation of the "magic" lists of constants {α j , β j } hard-coded in the snippet "4. Kosovo" is done by resorting to integer relation resolution in snippet "5. resolution". A formal computation of α j+2 , β j+2 from α j , β j is given in the next subsection. Remark 3. It is interesting to note that thanks to the inner symmetry within Kosovo, it is possible to determine α j+2 , β j+2 if α j , β j , α j+4 , β j+4 are known. Taking as an example j = 5, Figure 1 shows that Q 5,3,c = Q 3,3,c . Hence knowledge of α 3 , β 3 will be used to compute Q 3,3,c which is identical to Q 5,3,c .

Note that Q 5,5,c = Q 7,5,c . Hence knowledge of α 7 , β 7 will be used to compute Q 7,5,c which is identical to Q 5,5,c . Finally, having in hand Q 5,3,c , Q 5,5,c we can solve a system in two unknowns and determine α 5 , β 5 . Unfortunately this process has an information flow that only operates in "sandwich mode" allowing to determine α j+2 , β j+2 from α j , β j , α j+4 , β j+4 . This information flow cannot be reversed into an "escalator" allowing to ascend to level j + 4 from levels j and j + 2. The situation is illustrated in Figure 6 where x -→ y denotes the relation "y is computable from x". This limitation is solved at the next subsection where we infer α j+2 , β j+2 from α j , β j .

The Kosovo j-level master formula

Define:

(j) = 2 j j -1 2 ! 2 C j-3 2 α α j+2 = 4 α α j (j -4)(j -1)j(2j -5)(2j -3) -(3j -2) (j) (j -4)(j -1)(j + 1)
The formula is valid from j = 3 and on (for which

α α 3 = -1). β α j+2 = 4 β α j j(2j + 1)(j -1)(2j -5)(j -4) -3j 3 -17j 2 + 11j + 18 (j) (j -4)(j -3)(j + 1)
The formula is valid from j = 5 and on (for which

β α 5 = 234).
For j = 3 use directly

β α 3 = 4. α β j+2 = 4 j(j -6)(j -4)(j -2)(j -1)(2j -7)(2j -5) α β j -12 j 2 -4j + 2 (j) (j -6)(j -4) 2 (j -1)(j + 1)
The formula is valid from j = 3 and on (for which

α β 3 = -1/3).
The formula for β β j+2 is much more complex and does not fit in a single line. Define:

ϑ 1 (j) = (j -6)(j -4)(j -2)(j -1)j(2j -7)(2j -5)(-1 -3j + 2j 2 ) ϑ 2 (j) = 4(-390 + 312j + 653j 2 -942j 3 + 442j 4 -87j 5 + 6j 6 ) ϑ 3 (j) = (j -6)(j -4) 2 (j -1)(1 + j)(13 -11j + 2j 2 )
Then:

β β j+2 = 4 β β j ϑ 1 (j) -ϑ 2 (j) (j) ϑ 3 (j)
The formula is valid from j = 3 and on (for which

β β 3 = -14/3).
The test of those formulae proceeds in two steps. A first snippet ("16. files") computes the α j and β j for j = 3, 5, . . . , 501 using numerical simulation and records them in two files called file1.txt and file2.txt. Those files are read by snippet "17. j-level" that compares them to the values derived using the formal j-level formulae.

Remark 4. It is also possible to directly infer the α j,κ , β j,κ through symmetry. Note that the relation below also works for negative j, κ values.

Define:

ζ(j, u) = 1 2 u u-1 i=0 (2 + 2i -j) τ j,u =          ζ(j, u) |ζ(j, u)| • (2j -2u -3)!!(2u -j)!!(-2) u if 2u > j -1 ζ(j, u) • (-4) u otherwise
Then:

α j,j-u-1 α j-2u,j-u-1 = β j,j-u-1 β j-2u,j-u-1 = τ j,u
In particular τ j,j-1 = 1 2j-4 . The code is snippet "6. symmetry".

The Croatia Conjecture

As Bosnia & Herzegovina is a particular border case of Croatia the following is valid for both Bosnia & Herzegovina and Croatia.

Our automated software detected the following stunning behavior providing a κ-level formula for Croatia.

Let:

µ i,j = -(-2) 3j-11-4i 2 i q=1 (j -2q -2)
For every i = 0, 1, 2, . . . there exist two polynomials in j, denoted ψ 1 (i, j) and ψ 2 (i, j) such that for j ≥ 2i + 5 we have:

α j, j-2i-3 2 = ψ 1 (i, j) µ(i, j) and β j, j-2i-3 2 = ψ 2 (i, j) µ(i, j)
Tables 2 and3 provide the first values of the polynomials ψ 1 , ψ 2 . ψ 1 , ψ 2 can be computed algebraically because when (j, κ, c) ∈ Croatia, Q j,κ,c can be obtained by finite summation. We can hence derive α j, j-2i-3

2 and β j, j-2i-3 2 .
Then, by deriving enough (α j,• , β j,• ) pairs and knowing that deg j ψ 1 (i, j) = i and deg j ψ 2 (i, j) = i +1, we can compute ψ 1 (i, j) and ψ 2 (i, j) by interpolation. We did not code this tedious yet straightforward process.

See code snippet "7. Croatia". Fig. 6. Inferring α j+2 , β j+2 from α j , β j and α j+4 , β j+4 . The Q j,κ,c in green boxes are determined using integer relation algorithms.

i ψ 1 (i, j) 0 -1 1 14 -j 2 -464 + 58j -3j 2 3 27936 -4692j + 432j 2 -15j 3 4 -2659968 + 542256j -67836j 2 + 4260j 3 -105j 4 5 367568640 -86278560j + 13203480j 2 -1139700j 3 + 51450j 4 -945j 5 Table 2. ψ 1 (i, j) for 0 ≤ i ≤ 5. i ψ 2 (i, j) 0 -15 + 4j 1 306 -95j + 4j 2 2 -13360 + 4646j -357j 2 + 12j 3 3 999648 -379692j + 40368j 2 -2457j 3 + 60j 4 4 -113885568 + 46449360j -6124164j 2 + 513228j 3 -22935j 4 + 420j 5 5 18333538560 -7933530720j + 1224286440j 2 -126833100j 3 + 7864950j 4 -266175j 5 + 3780j 6 Table 3. ψ 2 (i, j) for 0 ≤ i ≤ 5.
Remark 5.

The leading coefficients of ψ 1 (i, j) (i.e. 1, 1, 3, 15, 105, 945, . . .) are (2i-1)!! whereas the leading coefficients of ψ 2 (i, j) (i.e. 4, 4, 12, 60, 420, 3780, . . .) are 4(2i -1)!!.

Remark 6. The ψ polynomials can always be written under a nested form, e.g.:

+ 10395 • (j -27))(j -25))(j -23))(j -21))(j -19))(j -17)
+ 41580 • (j -29))(j -27))(j -25))(j -23))(j -21))(j -19))(j -17)
Remark 7. The GCD between the u-th coefficient of ψ 1 (i, j) and the u-th coefficient of ψ 2 (i, j) is always smooth as illustrated in the code snippet "8. coefficients".

Balkans Knowledge Summary

In summary:

-Any Q j,κ,c value in Bosnia & Herzegovina is algebraically computable either by summation or by the Bosnian κ-level formula.

-In Croatia Q j,κ,c is directly computable by summation (hence making a jlevel formula superfluous) and, in addition, has κ-level formulae. -Any Q j,κ,c value in Montenegro is algebraically computable.

-For Kosovo we have j-level formulae.

-Serbia is fully determined by our knowledge of Montenegro and Kosovo.

-Negative j, κ, c values are of no interest as they provide finite summations.

We therefore have algebraic formulae for computing all Q j,κ,c values over all the Balkans. . The axes along which the κ-level master formulae operate in each area. We hence see that a finite amount of "magic" information in one dimension (that we fully provide in this paper) allows to algebraically compute Q j,κ,c over the two remaining dimensions.

Area j-level formula κ-level formula c-level formula

Croatia

not required Bosnia & Herzegovina not required Kosovo+Serbia Montenegro not required Table 4. Knowledge Summary A challenge, on which the authors are currently working, is characterizing the case of even j values (that yield formulae involving log 2).

As a motivational example let us unveil the simple log 2 example Q 2,κ,0 .

In this case:

Q 2,κ,0 = (-1) κ+1 a κ -b κ + a κ log 2
Where a κ is the LCM of the list of κ integers starting with κ (in Mathematica:

Table[Apply[LCM,Table[i,{i,k,2k-1}]],{k,1
,100]}) while the b κ are the numerators of the coefficients in the power series forlog(1 + x) log(1x).

Recall that the coefficients in the expansion of log(1 + x) log(1x) are given by 1 2

2n n

1 x=0 (x(1 -x)) n-1 log(x) dx and log(1 + x) log(1 -x) = 1 2 1 z=0 log(z) z(1 -z)       1 1 -4x 2 z(1 -z) -1       dz
Evidently, providing formal proofs of the formulae provided in this paper is a challenge by its own right. The Conservative Matrix Field approach of [6] is one promising direction to investigate. To date, attempts to code automated substitution-simplification-induction proofs were unsuccessful. Yet another interesting question is that of reversal : Given a Q j,κ,c (i.e. a 0 , a 1 , a 2 ) find j, κ, c [START_REF] Flórez | Characterization of the Strong Divisibility Property for Generalized Fibonacci Polynomials[END_REF] .

The main open questions remaining are very simple to formulate:

Open Question 1

Prove the formulae given in this paper.

Open Question 2

What happens in the Balkans for even j values?

Open Question 3

What happens in Inostranstvo (cf. remark 12)?

The following observations are hints that may serve in future quests:

Remark 8. Q 1,0,c = (2c)! 2(2c -1)!! 2 G -∆ c-1,0
where:

∆ 1,0,c =        1 + 10c if c < 2 2c(1 -2c) 3 ∆ 1,0,c-2 + (8c 2 + 2c + 1)∆ 1,0,c-1 if c ≥ 2
or under an equivalent more compact form:

∆ 1,0,c =        1 if c = 0 (2c)! + (2c + 1) 2 ∆ 1,0,c-1 if c > 0
and even [4]:

∆ 1,0,c-1 = (2c)!       2G 2c c 4 c - ∞ 0 t cosh 2c+1 (t) dt       Remark 9. lim c→∞ Q j,κ,c+1 -Q j,κ,c = 2 lim κ→∞ Q j,κ+1,c -Q j,κ,c = 2j lim κ→∞ Q j+1,j+2r+1,c -Q j,j+2r+1,c = 4r + 1
Remark 10. Denoting:

Q j,κ,c = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z

The following formula (code snippet "9. ratio") is valid all over areas:

ρ j,κ,c = j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1) (2c -2i + 1)(2κ -2i + 1) and ε j,κ = 2κ + j -7 2 + 1 j a 0 a 2 = (2c)! • 2 ε j,κ (2c -1)!! 2 • C κ-1 • C j-3 2 • (2κ -1) • (j -2) • ρ j,κ,c
Where ρ 1,κ,c = 1 by definition.

Remark 11. Although possibly unrelated, we note that low-degree continued fractions involving log 2 can be also obtained with lower degree polynomials, e.g. (See code snippet "10. log2-a".):

2 L( 1 2 , 1, c -1) = 2 ∞ n=0 e πin (n + c -1) = 1 2 c-2 log(2) - c-2 j=1 2 c-j-2 j = c + ∞ K n=1 -2n 2 3n + c
We get a similar behavior for:

R c = c + ∞ K n=1 -2n 2 -2n 3n + c
where 2 c-4 • (c -3) • a 0 = a 2 and for which we provide numerical examples in Table 12.

See code snippet "11. log2-b".

Remark 12. We also discovered other continued fractions involving G outside the Balkans. This suggests the existence of a more general formula encompassing both the Balkans and those other territories (called "Inostranstvo"). We denote those relations:

Q δ, ,τ,η,µ = + ∞ K n=1 -2n(n + τ)(n + η)(n + µ) + δn + 3n 2 = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z Luckily, given that we have two coefficients 10 in the denominator of the continued fractions, we can compare in one plot the coefficients and δ of the Balkans and of Inostranstvo in one figure (Figure 9). Similarly we can visualize in 3D the τ, η, µ of both regions (Figure 10). The alignments of red points show that there is clearly another structured family hiding out beyond the Balkans.

PCA and automated matching revealed that , δ are dependent on τ, η, µ and:

Q τ,η,µ = x + 2(τ + η + 1)i + ∞ K n=1 -2n(n + τ)(n + η)(n + 2i + µ) x + 2(τ + η + 1)i + (2(τ + η + µ + 2i) + 3)n + 3n 2 Where x = (1 + η)(1 + µ) + τ(1 + η + µ).
This formula works when all variables [START_REF] Lenstra | Factoring Polynomials with Rational Coefficients[END_REF] have identical parity, i.e.: For instance:

τ ≡ η ≡ µ mod 2
Q i = 7 + 6i + ∞ K n=1 -2n(n + 1) 2 (n + 2i + 1) 7 + 6i + (9 + 4i)n + 3n 2 for i = 0, 1 . . .
Whose formal expression (code snippet "12. Inostranstvo1") turns out to be:

Q i = (2i + 1)! ∆ i -2G(2i + 1)!! 2 ∆ i =        2 + 15i if i < 2 2(2i -1) 3 (1 -i) • ∆ i-2 + (8i 2 -2i + 3) • ∆ i-1 if i ≥ 2
or [START_REF] Naccache | On Catalan Constant Continued Fractions[END_REF] :

Q i = 23 + 10i + ∞ K n=1 -2n(n + 1)(n + 3)(n + 2i + 3) 23 + 10i + (17 + 4i)n + 3n 2 for i = 0, 1 . . .
We did not investigate further the various Inostranstvo families but conjecture that they share the same behaviors as the Balkans.

Remark 13. The ∆ functions appearing in this paper are particular cases of "Generalized Fibonacci Polynomials" (GFBs) studied by various authors, e.g. [START_REF] Naccache | On Catalan Constant Continued Fractions[END_REF] for which a 0 a 2 = (2j + 5)! (2j + 4)(2j + 5)!! 2 (code snippet "13. Inostranstvo2") Fig. 10. τ, η, µ for the Balkans (in blue) and Inostranstvo (in red). [9]. GFBs have numerous properties that might shed light on the open questions listed supra. We did not investigate this further.

Yet another route consists in considering the Inostranstvo and/or even-j targets as algebraic equations and attempting on them an algebraic sieving approach such as [3]. This was not investigated this process given its theoretical and logistical complexity.

Finally, the similarity between the infinite sums given in [13] and the continued fractions investigated in this paper may reveal connections allowing to prove our conjectures. While analyzing [13] we noted a probable misprint in the formulae given for y 3 and y 5 (bottom of page 9 of [13]). We hence conducted our own experiments and discovered the relations described in Table 5 where:

w 2 c + w 3 + w 1 ∞ n=1 (-1) n+1 7 i=1 (2n + 2i -3 + ) -e i = 0
12 The Gradient Descent Process

The Gradient Descent process that generated the formulae will be presented in steps. We sample execution at critical points using the example j = 11, κ = 6 and 40 ≤ c ≤ 47 to explain the automated exploration process.

The starting point

We first recall our notations:

Q j,κ,c = a 0 a 1 + a 2 G
where a 0 , a 1 , a 2 ∈ Z

We have by now "seen the end of the movie", and we know that for all areas:

n 0 (j, κ, c) = a 0 a 2 = (2c)! • 2 ε j,κ (2c -1)!! 2 • C κ-1 • C j-3 2 • (2κ -1) • (j -2) • ρ j,κ,c
Where:

ρ j,κ,c = j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1) (2c -2i + 1)(2κ -2i + 1) and ε j,κ = 2κ + j -7 2 + 1 j
We started our journey by manually inspecting n 0 (j, κ, c) for several j, κ, c values, noting that n 0 (j, κ, c) is always very smooth.

This suggests that n 0 (j, κ, c) is the product of basic combinatorial functions such as factorials, binomials, semifactorials, Catalan numbers, Pochhammer symbols etc.

But the question is -of coursewhich functions?

We now know that n 0 (j, κ, c) is an exotic zoo containing the following animals:

φ 0 = (2c)!, φ 1 = (2c -1)!!, φ 2 = (2c -1)!!, φ 3 = C κ-1 , φ 4 = C j-3 2 φ 5 = (2κ -1), φ 6 = (j -2), φ 7 = 2 2κ , φ 8 = 2 j-7 2 , φ 9 = 2 1 j φ 10 = j-1 2 i=1 (2c -2κ + 2i -1), φ 11 = j-1 2 i=1 (κ -i + 1), φ 12 = j-1 2 i=1 (2c -2i + 1)
φ 13 = j-1 2 i=1 (2κ -2i + 1) n 0 (j, κ, c) = φ 0 • φ 7 • φ 8 • φ 9 • φ 12 • φ 13 φ 1 • φ 2 • φ 3 • φ 4 • φ 5 • φ 6 • φ 10 • φ 11
However, at start, we had no idea what the φ i s were nor do we know how many φ i s are there.

Remark 14. The basic functions in our catalog are not independent as some multiplicatively generate others. e.g., Catalan numbers, binomials, multinomials and Pochhammer symbols are all products of factorials. Adding to the catalog 2 x we reach semifactorials etc. The code can hence successfully follow different paths for a given target n 0 (j, κ, c). While those functional dependencies do not impact the final result, they do impact complexity: e.g., using Catalan numbers reduces the depth of search [START_REF] Nimbran | Some Continued Fractions for π and G[END_REF] but increases its width.

Conversely, the code can also remove successfully components of φ i s and get stuck letter if the remaining (un-removed part) of the φ i is absent from the catalog.

Mutating functions

The algorithm performs a gradient descent on n i (j, κ, c), using an LLM to guide the descent. Catalog functions are not used in "bare metal" mode. They appear with specific linear combinations of j, κ, c, 1. To capture those combinations let:

σ ( ū) = u 0 j + u 1 κ + u 2 c + u 3 where ū = {u 0 , u 1 , u 2 , u 3 }
In other words [START_REF] Raayoni | Generating Conjectures on Fundamental Constants with the Ramanujan Machine[END_REF] :

φ 0 = σ (0, 0, 2, 0)!, φ 1 = φ 2 = σ (0, 0, 2, -1)!!, φ 3 = C σ (0,1,0,-1) φ 4 = C σ ( 1 2 ,0,0,-3 2 ) , φ 5 = σ (0, 2, 0, -1), φ 6 = σ (1, 0, 0, -2), φ 7 = 2 σ (0,2,0,0) φ 8 = 2 σ ( 1 2 ,0,0,-7 2 ) , φ 10 = σ ( 1 2 ,0,0,-1 2 )

i=1

(σ (0, -2, 2, -1) + 2i)

φ 11 = σ ( 1 2 ,0,0,-1 2 ) i=1 (σ (0, 1, 0, 1) -i), φ 12 = σ ( 1 2 ,0,0,-1 2 )
i=1 (σ (2, 0, 0, 1) -2i)

φ 13 = σ ( 1 2 ,0,0,-1 2 )
i=1 (σ (0, 2, 0, 1) -2i)

What information do we have?

The integer relations oracle [START_REF] Sanderson | א מ א ו ד נ י ע ס א ו נ ד ט ו ב . www.youtube.com/watch[END_REF] provides tens of thousands of n 0 (j, κ, c) instances.

We record these in a database B where each entry has the form: [START_REF] Nimbran | Some Continued Fractions for π and G[END_REF] When a Catalan number is identified three factorial identifications are avoided.

B i = {j i , κ i , c i , t i } = {j i , κ i , c i , n 0 (j i , κ i , c i )}
. The code needs to infer the ūs that intervene in n 0 (j, κ, c) from those numerous numerical examples.

A first step

Assume that the code is discovering the φ i s and their ūs one by one. The code is at some step ω at which it has already discovered the φ i s shown in red in the formulae:

n 0 (j, κ, c) = a 0 a 2 = (2c)! • 2 ε j,κ (2c -1)!! 2 • C κ-1 • C j-3 2 • (2κ -1) • (j -2) • ρ j,κ,c
And:

ρ j,κ,c = j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1) j-1 2 i=1 (2c -2i + 1)(2κ -2i + 1)
Divide the t i of each record B i by the red components evaluated at j i , κ i , c i . Update the target to:

n ω (j, κ, c) = (2c -1)!! 2 • C κ-1 • C j-3 2 • (2κ -1) • (j -2) • j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1)
A successfully mutated function, e.g. φ 1 = (2c -1)!! = σ (0, 0, 2, -1)!! standsout because:

n ω (j, κ, c) mod σ (0, 0, 2, -1)!! = 0 for all j, κ, c values Evidently, this criterion generates false positives. For instance:

n ω (j, κ, c) mod σ (0, 0, 2, -1)!! = 0 ⇒ n ω (j, κ, c) mod σ (0, 0, 2, -3)!! = 0
Remark 15. False positives come in two flavors: "False false positives" and "True false positives". σ (0, 0, 2, -3)!! is a "false false positive". Detecting σ (0, 0, 2, -3)!! is useful as it decreases the target but allows progress. In the example above, instead of peeling-off (2c -1)!! in one round, a first round will peel-off (2c -3)!! and leave an extra (2c -1) to some subsequent round. By opposition a "True false positive" is a candidate appearing as a factor of the target by the sole effect of chance over the available dataset.

To visualize efficiently execution we introduce Backgammon diagrams.

Backgammon diagrams

The gradient descent monitoring tool is called "Backgammon diagrams" because of its visual similarity to a backgammon board (Figure 11).

In a Backgammon diagram the x-axis shows a search performed over a coordinate u s within an interval u s ∈ {u start , . . . , u end }.

In all the following, consider that all ū coordinates other than s were fixed to correct values denoted by . This view is grandly simplified with respect to reality. In the code j, κ vary as well (resulting in multi-dimensional and hence unvisualizable diagrams) and several u s s are simultaneously tried at each round.

A worked-out example

We want to disassemble:

n ω (j, κ, c) = (2c -1)!! 2 • C κ-1 • C j-3 2 • (2κ -1) • (j -2) • j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1) step ω + 1 target n ω (j, κ, c) candidate (uc -1)!! ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 • • • • • • • • new target n ω+1 (j, κ, c) = n ω (j, κ, c)/(2c -1)!! database update ∀i do t i = t i /(2c i -1)!! remarks
We have one solution which is u = 2.

step illustrating option 1: ω + 2 target n ω+1 (j, κ, c) candidate illustrating option 1: (uc -1)!! ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+2 (j, κ, c) = n ω+1 (j, κ, c)/(2c -1)!! database update ∀i do t i = t i /(2c i -1)!! remarks
We have one solution which is u = 2.

step illustrating option 2: ω + 2 target n ω+1 (j, κ, c) candidate illustrating option 2:

(2c + u)!! ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5
new target n ω+2 (j, κ, c) = n ω+1 (j, κ, c)/(2c -1)!! database update ∀i do t i = t i /(2c i -1)!! remarks 4 solutions added to backtracking list.

step illustrating option 1: ω + 3 target n ω+2 (j, κ, c) candidate illustrating option 1: (uc -1)!! • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0
▲ ▲ ▲ ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+3 (j, κ, c) = n ω+2 (j, κ, c)/(2κ -
▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+3 (j, κ, c) = n ω+2 (j, κ, c)/(2κ -
▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0
▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+4 (j, κ, c) = n ω+3 (j, κ, c)/(j -
C κ+u ▲ ▲ ▲ ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+5 (j, κ, c) = n ω+4 (j, κ, c)/C κ-1 database update ∀i do t i = t i /C κ i -1 remarks
7 solutions added to backtracking list.

step illustrating option 2: ω + 5 target n ω+4 (j, κ, c) candidate illustrating option 2: C uκ-1 ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+5 (j, κ, c) = n ω+4 (j, κ, c)/C κ-1 database update ∀i do t i = t i /C κ i -1 remarks
2 solutions added to backtracking list.

step illustrating option 1: ω + 6 target n ω+5 (j, κ, c) candidate illustrating option 1:

j-1 2 x=1 (2c + uk + 2x -1) ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+6 (j, κ, c) = n ω+5 (j, κ, c)/ j-1 2 x=1 (2c -2k + 2x -1) database update ∀i do t i = t i / j i -1 2 x=1 (2c i -2k i + 2x -1)

remarks

We have one solution which is u = -2.

step illustrating option 2: ω + 6 target n ω+5 (j, κ, c) candidate illustrating option 2:

j-1 2 x=1 (2c -2k + 2x + u) ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+6 (j, κ, c) = n ω+5 (j, κ, c)/ j-1 2 x=1 (2c -2k + 2x -1) database update ∀i do t i = t i / j i -1 2 x=1 (2c i -2k i + 2x -1)

remarks

We have one solution which is u = -1.

step illustrating option 3: ω + 6 target n ω+5 (j, κ, c) candidate illustrating option 3:

j-1 2 x=1 (uc -2k + 2x -1) ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+6 (j, κ, c) = n ω+5 (j, κ, c)/ j-1 2 x=1 (uc -2k + 2x -1) database update ∀i do t i = t i / j i -1 2 x=1 (uc i -2k i + 2x -1)

remarks

We have one solution which is u = 2.

step illustrating option 4: ω + 6 target n ω+5 (j, κ, c)

candidate illustrating option 4: 2uj+j-1 2 x=1 (2c -2κ + 2x -1) ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 0 2 4 6 8 new target n ω+6 (j, κ, c) = n ω+5 (j, κ, c)/ 2uj+j-1 2 x=1 (2c -2κ + 2x -1) database update ∀i do t i = t i / 2uj i +j i -1 2 x=1 (2c i -2κ i + 2x -1)

remarks

We have one solution which is u = 0.

step

ω + 7 target n ω+6 (j, κ, c) candidate C j+2u+1 2 ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -3 -2 -1 0 1 2 3 new target n ω+7 (j, κ, c) = n ω+6 (j, κ, c)/C j-3 2 database update ∀i do t i = t i /C j i -3
2 remarks 3 solutions added to backtracking list. (κ + ux + 1)

step ω + 8 target n ω+7 (j, κ, c) candidate j-1 2 x=1 (κ -x + u) ▲ ▲ ▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+8 (j, κ, c) = n ω+7 (j, κ, c)/ j-1 2 x=1 (κ -x + 1) database update ∀i do t i = t i / j i -1 2 x=1 (κ i -x + 1)
▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -5 0 5 new target n ω+8 (j, κ, c) = n ω+7 (j, κ, c)/ j-1 2 x=1 (κ -x + 1) database update ∀i do t i = t i / j i -1 2 x=1 (κ i -x + 1)
remarks 2 solutions added to backtracking list. At this point one of the backtracking branches gives the constant function 1. n ω (j, κ, c) was hence disassembled.

The Pathfinder

As we have just seen, at any step the algorithm can take several paths each of which offering a different backtracking fan-out. At first we considered resorting to a 2D-backtracking where one parameter is the offered fan-out and the second is the φ-backtracking per se. The bookkeeping associated to this procedure seemed prohibitive, therefore we just opted to take the candidate with the least fan-out at each step. This appeared sufficient for our purpose and compatible with the computational means at hand.

The Decimator

A very significant speed-up is achieved by a software module called "the Decimator". The Decimator restricts the ū space by removing u i affine combinations incompatible with the target.

Consider the target:

n ω+2 (j, κ, c) = C κ-1 • C j-3 2 • (2κ -1) • (j -2) • j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1)
Create the database B given in Table 6. Assume that we want to test a candidate y(j, κ, c) = u 0 j + u 1 κ + u 2 c + u 3 . We are typically interested in exploring each u i over a small interval, e.g. [-8, 8]. Start by creating a list L of all (2 × 8 + 1) 4 u-values.

For each {u 0 , u 1 , u 2 , u 3 } value, if a t i is not divisible by at least one y(j i , κ i , c i ) then remove {u 0 , u 1 , u 2 , u 3 } from L.

Table 7 When we merge all the forbidden us, removing duplicates, we get a collection of 78567 combinations to skip. We are hence left with (2 × 8 + 1) 4 -78567 = 4954 survivors to test, i.e. 5.9%.

The above example is restricted to eight c toy-values. In reality we perform calculations on 10 5 combinations of (j, κ, c). This reduces drastically the search space 17 . Note however that as c increases the percentage of newly removed ū values per round decreases.

Note that, in practice even an exploration in [-3, 3] 4 would have sufficed 18 . For u ∈ [-3, 3] 4 we get 332 survivors out of 2401 (14%).

For [-8, 8] 4 and 40 ≤ c ≤ 47, decimating for y(j, κ, c) = C u 0 j+u 1 κ+u 2 c+u 3 removes 83233 candidates, leaving only 288 possibilities (0.35%).

Decimating over [-5, 5] 5 and 20 ≤ c ≤ 60 for the candidate:

y(j, κ, c) = j-1 2 i=1 (u 0 j + u 1 κ + u 2 c + u 3 + u 4 i)
results in 1496 survivors out of 161051 (0.93%).

Counter-intuitively, the more "complex"19 the candidate is the more efficiently it is decimated. We hence select the candidates in decreasing complexity order. Measuring the complexity of mathematical formulas can be a subjective task, as it depends on various factors such as the number of terms, the presence of functions, exponents, and variables, as well as the overall structure. There's no definitive metric to quantify formula complexity. The following criteria can nonetheless be used to "measure" complexity and hence get rid of φ i s as fast as possible:

Counting Elements: We can count the number of distinct elements in each candidate, such as variables, constants, operators, and functions. For example, the candidate κ -1 has two elements while in others we may have multiple variables, exponents, a product symbol, and a summation, which increases its complexity.

Nesting and Hierarchy: Analyze the nesting of operations and functions within the candidates. A candidate with multiple levels of nesting or hierarchy can be regarded as more complex.

Mathematical Operations: Consider the types of mathematical operations present in the candidates. More complex operations, such as exponentiation and summation, contribute to higher complexity compared to simpler operations like addition or multiplication.

Function Complexity: If the candidate includes functions, their complexity should be taken into account. For instance, a Catalan number can add complexity compared to linear or constant functions.

Symbolic Representation: Represent each candidate in a symbolic format, such as a parse tree or abstract syntax tree. Compare the depth and branching of the trees as a rough measure of complexity.

Information Theory: Explore concepts from information theory, such as Kolmogorov complexity or algorithmic information theory, to quantify the amount of information needed to describe each candidate. This approach can be quite theoretical and may not provide a practical measure for all cases.

We resorted to a much more brutal approach [2]. Using the API20 , we asked GPT-4 to compare candidates pairwise, obtained a subjective complexity comparison (≺) of each pair and translated the ternary results21 to a directed graph. Because the LLM does not provide consistent answers (i.e. it might say that A ≺ B ≺ C and. . . C ≺ A) we performed a random walk of 10 6 steps on the graph and counted the number of times each candidate was visited. The most visited candidate was considered as the "most complex". We then removed this candidate from the graph and started over again.

As a final note, we remark that some φ i s are "process killers". This is the case of candidates such as (2c)!. We called φ i s silencers "opioids" are they efficiently remove the symptoms but do not remove any lurgy. Because, in essence, (2c)! contains just about any number of interest, it silences the modular tests. We hence start by launching the process with ū! and start the backtracking afresh for each n 0 (j, κ, c)/ ū!. Luckily, the early investigation of Bosnia & Herzegovina by which we started allowed to uncover the opioids before formulae get too complex.

Ascending to descend

As underlined in [15], some descents require intermediate ascents. The process described so far advances only in the case of a monotonous descent. In other words, if the initial φ i s were wrongly chosen the process will not converge.

We thus need a process allowing temporary ascents to get out of bowls. To that end we use a modified version of Broyden-Fletcher-Goldfarb-Shanno's (BFGS) algorithm. This requires a more refined measure of the penalty/profit of each move which cannot just be the Boolean "x divides y". Denote by p i the i-th prime. We introduce a measure called "brittleness"22 denoted Ξ(n).

Let: n = a-1 i=0 p m i i ∈ Q be a simplified fraction. Then Ξ(n) = a-1 i=0 |m i |
In other words, Ξ(n) counts, with repetition, the number of distinct factors appearing in either the numerator or the denominator of n.

The following example illustrates the evolution of brittleness (y-axis) during the peeling-off process. Dotted lines ⇒ Target's brittleness at the concerned j, κ, c points. Draughts ⇒ Raised to Ξ(target/candidate) for different u values. As before, denotes us at which all draughts are lower than their samecolor dotted lines. The thin red vertical line is the correct answer.

As an example start with the target:

n 1 (j, κ, c) = C κ-1 • C j-3 2 • (2κ -1) • (j -2) • j-1 2 i=1 (2c -2κ + 2i -1)(κ -i + 1)
Figure 13 represents the following steps: ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲▲ ▲ ▲▲▲ ▲▲ ▲ ▲▲ ▲ ▲▲▲ ▲ ▲▲▲▲▲ ▲ ▲ ▲ ▲▲

step 1 of Figure 13 target n 0 (j, κ, c) candidate illustrating first option: uκ -1 new target n 1 (j, κ, c) = n 0 (j, κ, c)/(2κ -1)
new target n 3 (j, κ, c) = n 2 (j, κ, c)/ j-1 2 x=1 (2c -2k + 2x -1) ▲ ▲ ▲▲ ▲ ▲ • ••• • •• • • •• • • • • •• • ••• • •• • • •• ••• • •• • ••• • •• • • •• ••• • •• • ••• • •• • • •• ••• • •• • ••• • •• • • ••• •• • •• • ••• • •• • • ••• • • • •• • ••• • •• • • ••• • • • •• • ••• • • •• • ••• • • • •• -5 0
• • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • • • • • • •• • • • •• • • • • • -5 0
• • • • •• • • • • • • • • • • • • •• •• • • • • • • • • • • • • • •• •• • • • • • • • • • • • • • •• •• • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • •• • • •• • • • • • • • • -5 0
• • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • • •• • • • • •• •• • • • • • • •• • • • • • -5 0 5 0 5
▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ A Tables of Constants Note that the Montenegro line does not obey that same c-level formula but was validated by snippet "1. Montenegro". Note that the formula is also tested for negative j values (not shown as red dots in the diagram).

••• • • • • • • • •• • •• • • ••• • • • • • • •• • •• • • • ••• • • • • • ••• ••• • • • ••• • • • • ••• ••• • • • • ••• • • • • •• •••• • • • • ••• • • • • • •••• • • • • • ••• • • • • •••• • • • • • • ••• • • • • ••• • • • • • • • -5 0
▲ ▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 0 2 4
▲▲▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -2 0
▲▲▲▲▲▲▲▲▲ • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • -2 0
(1 + n)(3 + n)(13 + n) + 35n -28028 -999391 1441440 (2 + n)(7 + n)(13 + n) 112 + 43n -27720 -20417 27720 n(1 + n)(11 + n) + 27n -22050 27649 -40320 (3 + n)(7 + n)(9 + n) + 35n -19305 424423 -612612 (1 + n)(5 + n)(15 + n) + 43n -17640 -250007 360360 n(7 + n)(9 + n) + 35n -14700 -153907 221760 (2 + n)(7 + n)(9 + n) + 35n -10395 124741 -180180 (1 + n)(5 + n)(11 + n) + 35n -9450 76691 -110880 (1 + n)(5 + n)(9 + n) +
(1 + n)(5 + n) 2 + 23n -350 1739 -2520 (1 + n)(5 + n)(7 + n) + 27n -180 -299 420 n(3 + n)(5 + n) + 19n -105 142 -210 (1 + n)(3 + n)(7 + n) + 23n -18 7 -12 (1 + n)(3 + n) 2 + 15n -6 -5 6 n(1 + n)(3 + n) 8 + 11n
(1 + n)(5 + n) + 15n 90 -79 120 (1 + n)(3 + n)(5 + n) + 19n 420 319 -420 n(1 + n)(7 + n) + 19n 420 887 -1260 n(3 + n)(7 + n) + 23n 900 361 -480 (2 + n)(5 + n) 2 + 23n 1890 -3443 5040 (1 + n)(3 + n)(9 + n) + 27n 2100 2377 -3360 (2 + n)(5 + n)(7 + n) + 27n 5544 50035 -72072 n(5 + n)(11 + n) + 35n 6468 -249713 360360 (1 + n)(7 + n)(11 + n) + 39n 7560 19409 -27720 n(3 + n)(9 + n) + 27n 8316 -19031 27720 (1 + n)(3 + n)(11 + n) + 31n 15444 -49705 72072 (1 + n)(3 + n)(15 + n) + 39n 22050 -499279 720720 (1 + n)(7 + n)(9 + n) + 35n 24255 -76586 110880 (3 + n)(7 + n)(11 + n) + 39n 25740 200107 -288288 (2 + n)(5 + n)(15 + n) + 43n 37800
P (n)/(-2n) T (n) -3n 2 -80281600 -10675439 9459450 (2 + n)(3 + n)(14 + n) + 35n -15728640 392683 -727650 (4 + n) 2 (12 + n) + 35n -13107200 -263867 -28350 (4 + n)(5 + n)(10 + n) + 31n -10485760 -93699 -103950 (4 + n)(5 + n)(12 + n) + 35n -7372800 -884203 727650 (2 + n)(3 + n)(12 + n) + 31n -6291456 149419 -257250 (4 + n)(8 + n) 2 + 35n -5242880 -86807 9450 (5 + n)(6 + n)(10 + n) + 35n -3932160 -116317 3150 (4 + n)(5 + n)(8 + n) + 27n -3276800 -158859 22050 (2 + n)(5 + n)(10 + n) + 27n -2621440 -48609 -1050 (5 + n)(6 + n)(8 + n) + 31n -2359296 -168445 103950 (2 + n)(4 + n)(12 + n) + 31n -491520 50593 -66150 (3 + n)(4 + n)(10 + n) + 31n -327680 -21271 9450 (2 + n)(4 + n)(10 + n) + 27n -230400 -1909 -22050 (2 + n) 2 (5 + n) 9 + 11n -196608 -184547 198450 (3 + n)(6 + n)(10 + n) + 35n -163840 -4981 -22050 n(5 + n)(8 + n) 9 +
(2 + n)(4 + n) 2 + 15n -2048 -43 -30 (3 + n)(4 + n)(6 + n) + 23n -768 -77 -18 (2 + n)(3 + n)(4 + n) + 15n -288 31 -90 n(2 + n)(3 + n) 3 + 7n
)(5 + n)(8 + n) + 23n 9830400 -1833409 2182950 (3 + n)(4 + n)(12 + n) + 35n 12582912 184025 -7350 (5 + n)(8 + n) 2 + 35n 39321600 1965547 -727650 (2 + n)(5 + n)(12 + n) + 31n 165150720 12969199 -9459450 (2 + n)(4 + n)(14 + n) + 35n 330301440 17687791 -9459450 (2 + n)(5 + n)(14 + n) + 35n
C.4 If 3 ≤ j ≤ κ + 2 3 ≤ j ≤ κ + 2 is in Kosovo. In which case we proceed in three steps:

Step 1: Define: The formula is valid from j = 3 and on (for which α α 3 = -1).

β α j+2 = 4 β α j j(2j + 1)(j -1)(2j -5)(j -4) -3j 3 -17j 2 + 11j + 18 (j) (j -4)(j -3)(j + 1)

The formula is valid from j = 5 and on (for which The formula is valid from j = 3 and on (for which α β 3 = -1/3).

The formula for β β j+2 is: Define: ϑ 1 (j) = (j -6)(j -4)(j -2)(j -1)j(2j -7)(2j -5)(-1 -3j + 2j 2 ) ϑ 2 (j) = 4(-390 + 312j + 653j 2 -942j 3 + 442j 4 -87j 5 + 6j 6 ) ϑ 3 (j) = (j -6)(j -4) 2 (j -1)(1 + j)(13 -11j + 2j 2 ) Then:

β β j+2 = 4 β β j ϑ 1 (j) -ϑ 2 (j) (j) ϑ 3 (j)
The formula is valid from j = 3 and on (for which

  k = 2 k = 3 k = 4 k = 5 k = 6 k = 7 k = 8 k = 9 k = 10 k = 11 k = 12 k = 13 k = 14

Fig. 1 .

 1 Fig. 1. The Five j, k Areas. The meaning of the arches connecting Kosovo, Serbia, and Montenegro will be clarified later.

Fig. 2 .

 2 Fig. 2. The areas of Figure 1, with the dimension c added. Each point in space corresponds to a Q j,κ,c value.

Q 1 ,

 1 κ,c = 2κ + 1 + ∞ K n=1 -2n 2 (n + 2κ)(n + c) 3n 2 + (3 + 4κ)n + 2κ + 1 Define the functions:

Fig. 4 .

 4 Fig. 4. j, κ and c-level master formulae. This paper provides exact algabraic processes for inferring Q j,κ,c for all areas.

ψ 1 (

 1 j, 6) = -14487726825 -(104826150 + (452605725 + (121200300 + (13697775 + (640710

ψ 2 (

 2 j, 6) =3198013886925 + (145296572850 + (5207427225 + (4353102000 + (877052475 + (78210090 + (3023055

Fig. 7 .

 7 Fig. 7. Q j,κ,c Requiescat in pace.

Fig. 8

 8 Fig.8. The axes along which the κ-level master formulae operate in each area. We hence see that a finite amount of "magic" information in one dimension (that we fully provide in this paper) allows to algebraically compute Q j,κ,c over the two remaining dimensions.

Fig. 9 .

 9 Fig. 9. , δ for the Balkans (in blue) and Inostranstvo (in red).

Fig. 11 .

 11 Fig. 11. A typical Backgammon board.

  The draughts represent experiments with fixed j, κ values and different c values. In the example j = 11, κ = 6 and 40 ≤ c ≤ 47. Each draught color corresponds to a different c value (•,•,•,•,•,•,•,•). The legend is not repeated to save space. Draughts were slightly lifted to avoid covering each other. If for a given c value: n ω (j, κ, c) mod φ i (σ ( , . . . , , u s , , . . . , )) = 0 then the draught of c's color is lowered to the bottom of diagram, else it it raised to the top. Hence, a glance at the diagram shows which u s values are compatible with the target n ω (j, κ, c) for each c. Two additional features enhance reading: a red line showing the correct answer 16 and little red triangles ( ) denoting u s values for which tests succeeded for all c values.

Fig. 12 .

 12 Fig. 12. Q j,κ,c Process killers.

step 1 of Figure 13 target n 0 (

 0 j, κ, c) candidate illustrating second option: 2κ + u new target n 1 (j, κ, c) = n 0 (j, κ, c)/(2κ -1) step 2 of Figure 13 target n 1 (j, κ, c) candidate uj -2 new target n 2 (j, κ, c) = n 1 (j, κ, c)/(j -2k + ux -1)new target n 3 (j, κ, c) = n 2 (j, κ, c)/

Fig. 13 .1 2 3 3 4 5 6 . 1 step 5 2 step 6

 1361526 Fig.13. Steps are ordered from left→right and up→down, i.e. 1 1 2 3 3 4 5 6 . Note the steady decrease of Ξ levels. Steps 1 and 1 are two division options occurring at the same step and so are 3 and 3 .

  values and the first β j = { α β j , β β j } values.B Mathematica Code( * code snippet "1. Montenegro" * ) NumQ[k _ ,c _ ]:=2k+1+ContinuedFractionK[-2n^2(n+2k)(n+c),3n^2+(3+4k)n+2k+1,{n,1,5000}]; vD[a _ ,b _ ,k _ ,c _ ]:=If[c<2,a+b * c,-2c(2c-1)(2(c-k)-1)^2vD[a,b,k,c-2]+(8c^2+(2-8k)c-2k+1)vD[a,b,k,c-1]]; vG[a _ ,b _ ,k _ ,c _ ]:=((2c-1)!!)^2Catalan+Product[(2(c-j)-1),{j,0,k-1}] * vD[a,b,k,c-1]; vd[k _ ]:=4^(k-1)/(2k-1)/CatalanNumber[k-1];vr[k _ ]:=vd[k](-1)^(k)(1-2k)/((2k)!(2k-3)!!); va[k _ ]:=vr[k]vD[1,-2,1,k-1];vb[k _ ]:=-vr[k](2k-3)^2vD[1,12,2,k-1]-va[k]; QFor[k _ ,c _ ]:=vd[k](2c)!/vG[va[k],vb[k],k,c]; Print[Union[Flatten[Table[N[QFor[k,c],200]==N[NumQ[k,c],200],{c,1,14},{k,1,14}]]][[1]]];ClearAll["Global' * "];

Fig. 14 .

 14 Fig. 14. The points on which Montenegro is tested for 1 ≤ c ≤ 14 by the snippet "1. Montenegro" are shown in red.

Fig. 15 .

 15 Fig. 15. The points on which Bosnia & Herzegovina is tested for 1 ≤ c ≤ 14 by the snippet "2. Bosnia" are shown in red.

( * code snippet " 3 .

 3 Northern Balkans" * ) NumQ[j _ ,k _ ,c _ ]:=j(2-j+2k)+ContinuedFractionK[-2n(c+n)(j+n-1)(1-j+2k+n),3n^2+(3+4k)n+j(2-j+2k),{n,1,15000}]; vD[ab _ ,j _ ,k _ ,c _ ]:=If[c<2,ab[[1]]+ab[[2]]c,-2c(2c-j)(2c-2k+j-2)(2c-2k-1)vD[ab,j,k,c-2]+(8c^2+(2-8k)c+(j-2)(2k-j))vD[ab,j ,k,c-1]]; f[j _ ,k _ ,c _ ]:=Product[(2c-2k+2i-1)(k-i+1),{i,1,(j-1)/2}]CatalanNumber[(j-3)/2](j-2)(2k-1) * (2c-1)!!^2 * CatalanNumber[k-1]; g[j _ ,k _ ,c _ ]:=Product[(2c-2i+1)(2k-2i+1),{i,1,(j-1)/2}](2c)!2^(2k+(j-7)/2); h[j _ ,k _ ,c _ ]:=Product[2c-2i-1,{i,0,(j-3)/2}]Product[2c-2i-1,{i,0,k-1}]; Q[j _ ,k _ ,c _ ,ab _ ]:=Simplify[g[j,k,c]/(vD[ab,j,k,c-1]h[j,k,c]+f[j,k,c]Catalan)]; Ratio:=Function[{j,k,c},r=N[NumQ[j,k,c],2000];v=FindIntegerNullVector[{1,r,N[Catalan * r,2000]}];-v[[1]]/v[[2]]]; GenAB[j _ ]:=Table[({a,b}/.Solve[Table[{a,b}.{1,c-1}==g[j,k,c]/Ratio[j,k,c]/h[j,k,c],{c,1,2}],{a,b}])[[1]],{k,1,6}]; result={};For[j=-7,j<=13,If[j==1,j=3];AB=GenAB[j];AppendTo[result,Union[Flatten[Table[N[Q[j,k,c,AB[[k]]]==NumQ[j,k,c ],20],{k,1,6},{c,1,7}]]]];j+=2];Print[Union[Flatten[result]][[1]]];ClearAll["Global' * "];

Fig. 16 .

 16 Fig.16. The points on which the c-level master formula is tested for 1 ≤ c ≤ 7 by the snippet "3. Northern Balkans" are shown in red. Note that the Montenegro line does not obey that same c-level formula but was validated by snippet "1. Montenegro". Note that the formula is also tested for negative j values (not shown as red dots in the diagram).

  j -4)(j -1)j(2j -5)(2j -3) -(3j -2) (j) (j -4)(j -1)(j + 1)

  j -6)(j -4)(j -2)(j -1)(2j -7)(2j -5) α β j -12 j 2 -4j + 2 (j) (j -6)(j -4) 2 (j -1)(j + 1)

β β 3 =Step 3 : 2 C κ- 1 2 j-1 2 i=1(

 332122 -14/3).Using those formulae, iterate on j to inferα i)(2κ -2i -1) 2 (n, j, κ) = (-1) κ+1 (2κ)! 2 κ!2 3κ-2 (2κj)(2κ -1)(n((2κj -2)(3 -2κ) -1) + 1) • π(j, κ) η(n, j, κ) = (2κ + 2j -9 -2n)(2κ + j -8 -2n)(-2κ + 5j)(2κ + j -6) φ(n, j, κ) = 8κ 2 + κ(10j -48 -8n) + 3j 2 -(28 + 4n)j + 68 + 18n ∆n,j,κ (α, β) = if κ < 2 η(n, j, κ) • ∆n,j,κ-2 (α, β) + φ(n, j, κ) • ∆n,j,κ-1 (α, β) if κ ≥ 2 Define: ∆ j,κ,c (α j,κ , β j,κ ) = κ + β j,κ c if c < 2 -2c(2cj)(2c -2κ + j -2)(2c -2κ -1)∆ j,κ,c-2 (α j,κ , β j,κ ) if c ≥ 2 + (8c 2 + (2 -8κ)c + (j -2)(2κj))∆ j,κ,c-1 (α j,κ , β j,κ ) f j,κ,c = C j-3 (j -2)(2κ -1)(2c -1)!! 2c -2κ + 2i -1)(κi + 1)g j,κ,c = (2c)!2 κ,c = g(j, κ, c) ∆ j,κ,c-1 (α j,κ , β j,κ ) • h(j, κ, c) + f (j, κ, c) • G

  j+4,κ 2 and β j+4,κ 2 α j,κ 4 and β j,κ 4 Q j+4,κ 2 ,c 3 and Q j+4,κ 2 ,c 4 Q j,κ 4 ,c 7 and Q j,κ 4 ,c 8 α j+4,κ 1 and β j+4,κ 1 α j,κ 3 and β j,κ 3 Q j+4,κ 1 ,c 1 and Q j+4,κ 1 ,c 2 Q j,κ 3 ,c 5 and Q j,κ 3 ,c 6

α j+4 and β j+4 α j and β j ∧ ∧ α α j+4,j+2 and β j+4,j+2 α j,j and β j,j Q j+4,j+2,c 9 and Q j+4,j+2,c 10 Q j,j,c 11 and Q j,j,c 12 Q j+2,j+2,c 9 and Q j+2,j+2,c 10 Q j+2,j,c 11 and Q j+2,j,c 12 α j+2,j+2 and β j+2,j+2 α j+2,j and β j+2,j ∧ α j+2 and β j+2 any α j+2,κ and β j+2,κ any Q j+2,κ,c

Table 5 .

 5 Relations for π, G and log 2. See code snippet "15. Series".

		c	e 1 , e 2 , . . . , e 7	w 1 , w 2 , w 3
	0	π	0, 0, 2, 2, 2, 2, 0 57153600, -33075, 103904
	0	π	0, 1, 1, 2, 2, 1, 1 69854400, -24255, 76192
	0	π	0, 1, 2, 1, 1, 2, 1 62868960, -14553, 45712
	0	π	0, 2, 2, 1, 2, 2, 0 -65318400, -4725, 14848
	0	π	0, 2, 2, 2, 2, 0, 0	-6350400, -3675, 11552
	0	π	1, 1, 2, 2, 1, 1, 0	-907200, -315, 992
	0	π	1, 2, 1, 1, 2, 1, 0	-635040, -147, 464
	0	π	1, 2, 2, 2, 2, 1, 0	16934400, -245, 768
	0	π	2, 1, 1, 1, 1, 2, 0 -59535000, -6615, 21292
	0	π	2, 2, 0, 0, 2, 1, 1	-93139200, -2695, 9344
	0	π	2, 2, 0, 0, 2, 2, 0	-52920000, -2695, 9056
	0	π	2, 2, 1, 2, 2, 0, 0 -50803200, 3675, -11264
	0	π	2, 2, 2, 2, 0, 0, 0	129600, -75, 224
	0	G	0, 1, 2, 2, 2, 1, 0 -50803200, 66150, -60577
	0	G	0, 3, 2, 3, 0, 0, 0	-3456000, -6750, 6197
	0	G	1, 2, 2, 2, 1, 0, 0	-2419200, -3150, 2909
	0	G	1, 3, 0, 3, 1, 0, 0	8064000, 8750, -8109
	0	G	2, 2, 0, 2, 2, 0, 0 -33868800, -22050, 21131
	0	G	3, 2, 3, 0, 0, 0, 0	-27648, 54, -25
	1 log 2 0, 0, 0, 2, 2, 2, 2 -33177600, -38400, 26617
	1 log 2 0, 0, 2, 2, 2, 2, 0	1382400, -1600, 1109
	1 log 2 0, 1, 1, 2, 2, 1, 1	11059200, -7680, 5323
	1 log 2 0, 1, 2, 1, 1, 2, 1 -22118400, 10240, -7097
	1 log 2 0, 2, 2, 0, 0, 2, 2	17280000, -1760, 1219
	1 log 2 0, 2, 2, 2, 2, 0, 0	442368, 512, -355
	1 log 2 1, 1, 2, 0, 0, 2, 2	-88473600, 5120, -3539
	1 log 2 1, 1, 2, 2, 1, 1, 0	-230400, -160, 111
	1 log 2 1, 2, 1, 1, 2, 1, 0 -22118400, -10240, 7109
	1 log 2 2, 2, 0, 0, 2, 1, 1	-88473600, -5120, 3627
	1 log 2 2, 2, 0, 0, 2, 2, 0	69120000, 7040, -4951
	1 log 2 2, 2, 1, 2, 2, 0, 0	7077888, -1024, 707
	1 log 2 2, 2, 2, 2, 0, 0, 0	-13824, 16, -11

Table 6 .

 6 The database B

	i	j i	κ i	c i	t i = n ω+2 (j i , κ i , c i )
	1	11	6	40	86562004597992000
	2	11	6	41	99107222655672000
	3	11	6	42	113065986409992000
	4	11	6	43	128554477699032000
	5	11	6	44	145695074725569600
	6	11	6	45	164616513001617600
	7	11	6	46	185454046292961600
	8	11	6	47	208349607563697600

Table 7 .

 7 gives the number of us eliminated from [-8, 8] 4 by each c value. Number of eliminated u combinations per c tried.

	c	40	41	42	43	44	45	46	47
	eliminated us 51721 54371 55635 56161 52771 51203 51609 46383

Table 8 .

 8 Examples of convergence to

	a 0
	a 1 +a 2 log 2

Table 9 .

 9 Examples of convergence to a 0 a 1 +a 2 log 2 . The entry in blue is the one reported by the Ramanujan Project.

	a 0	a 1	a 2

Table 10 .

 10 Examples of convergence to

	a 0
	a 1 +a 2 G

Table 11 .

 11 Examples of convergence to

	a 0
	a 1 +a 2 G

Table 12 .

 12 Examples of convergence to

	a 0 a 1 +a 2 log 2 for R c

Table 13 .

 13 The first α j = {

This symmetry simply comes from the fact that the equation j(2 -2j + 2κ) = x(2 -2x + 2κ) has the two solutions x = j and x = κj + 1.

i.e. using an integer relation algorithm.

This somewhat unusual notation is used to note the "α of α", the "β of α" etc., as the same type of formula is applied at both c and κ levels.

We exclude the 3 of 3n 2 in the denominator, which is common to both the Balkans and to Inostranstvo.

(code snippet "14. Inostranstvo")

 j was not in the catalog and was added manually to unify two families of general formulae discovered by two runs.

e.g., exploring for 20 ≤ c ≤ 60 reduces the survivors' pool to 3702 (4.4%).

i.e., closer to a random oracle.

endpoint https://api.openai.com/v1/engines/davinci-codex/completions

"A is more complex that B": A → B, "B is more complex than A": B → A or "unsure": no edge.

Brittleness is a generalization the prime Ω function to Q: Ξ(n) = Ω(numerator(n)) + Ω(denominator(n)).

( * code snippet "6. symmetry" * ) ( * code snippet "7. Croatia" * )

( * code snippet "9. ratio" * )

* v[[1]]==v[[3]],-v[[1]]/(v[[2]]+Log[2]v[[3]])}];c++];ClearAll["Global' * "];

( * code snippet "12. Inostranstvo1" * )

C Altogether

This appendix recaps the entire algorithm. It takes as input j, κ, c and returns the exact expression of Q j,κ,c where:

In this case (j, κ, c) ∈ Bosnia & Herzegovina ∪ Croatia, hence Q j,κ,c is computed by straightforward finite summation.

In this case we are in Serbia. We hence use the symmetry relation:

Replace j by j = 2(κ + 1)j and compute Q j ,κ,c using the formula for Montenegro or for Kosovo given in the next subsections.

C.3 If j = 1

In this case we are in Montenegro. We hence define:


Rodrigo Vicente Cruz

Eric Lamballais
email: eric.lamballais@univ-poitiers.fr

Rodolphe Perrin

Implicit Wall-Layer Modelling in Turbulent Pipe Flow

Introduction

In the context of Large Eddy Simulations (LES) of wall-bounded turbulence, the explicit calculation of all the energy containing near-wall eddies is a key point. To accomplish this goal, it is recommended the employment of a mesh resolution between 10 and 100 wall units for the streamwise direction discretization, from 10 to 20 wall units in the spanwise direction, and typically one wall unit in the wall normal direction through a near-wall mesh refinement. When these resolution criteria are too demanding in terms of computational resources, an alternative is to use a discretization that bypasses the inner layer with the aid of a wall-layer model [START_REF] Piomelli | Wall-layer models for large-eddy simulations[END_REF][START_REF] Sagaut | Large eddy simulation of incompressible flow: an introduction[END_REF].

However, it was shown by [START_REF] Dairay | Mesh node distribution in terms of wall distance for large-eddy simulation of wall-bounded flows[END_REF] that excellent basic turbulent statistics in a pipe flow can be obtained with a mesh resolution coarser than the viscous sublayer without using any explicit wall-layer modelling. In this context, the near-wall regularization is simply performed by means of an original and robust implicit LES (ILES) technique that -without any extra-cost or adaptation in the near-wall region -displays a wall modelling feature. Two Reynolds numbers Re τ = 180, 550 1 were considered when comparing ILES results to unfiltered DNS data of [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF].

The term implicit LES refers here to the use of regularization as a substitute of subgrid-scale (SGS) modelling [START_REF] Dairay | Numerical dissipation vs. subgrid-scale modelling for large eddy simulation[END_REF][START_REF] Lamballais | Vassilicos: Implicit/explicit spectral viscosity and large-scale SGS effects[END_REF]. The originality of the method is found in the built-in numerical dissipation that does not come from the discretization of the convective term, but from the viscous term in the Navier-Stokes equation. This strategy has also been used successfully by [START_REF] Lamballais | Straightforward high-order numerical dissipation via the viscous term for direct and large eddy simulation[END_REF][START_REF] Dairay | LES of a turbulent jet impinging on a heated wall using high-order numerical schemes[END_REF] for turbulent plane channel and impinging jet configurations with regular Cartesian meshes, nonetheless, the nearwall behaviour of the implicit SGS modelling was not specifically investigated.

Until now, the aforementioned conclusions about the ILES technique performance were based on comparisons of basic statistics with unfiltered DNS data, which means that even if a good agreement was observed, no rigorous assessment could be made. In the present investigation, this matter is examined at higher Reynolds number (Re τ = 1000) while determining to what extent it can be generalized in terms of near-wall region bypass. To do so, a quasi-DNS database is presented with an original method to easily produce filtered statistics in order to enable rigorous comparisons with ILES at any mesh resolution. The implicit LES results are, thus, analysed in terms of ability to predict near-wall statistics despite the use of very coarse grid for which even the near-wall turbulent production region is fully bypassed.

Numerical Method and Computational Configuration

As a generic numerical tool to investigate this concept of implicit LES, the sixthorder flow solver Incompact3d is used. Thanks to its features of kinetic energy conservation 2 and flexibility for the application of numerical dissipation, this code enables an easy calibration of the implicit SGS modelling. In particular, the role of its scale selectivity has been shown in [START_REF] Dairay | Numerical dissipation vs. subgrid-scale modelling for large eddy simulation[END_REF] through rigorous comparisons with DNS results based on a priori and a posteriori analysis with a particular attention to distant triad interactions between SGS and very large scales [START_REF] Lamballais | Vassilicos: Implicit/explicit spectral viscosity and large-scale SGS effects[END_REF]. In addition, it was shown by [START_REF] Dairay | Numerical dissipation vs. subgrid-scale modelling for large eddy simulation[END_REF] that the present implicit LES technique is equivalent to the use of spectral vanishing viscosity (SVV), ensuring high-order accuracy and flexible application. In previous studies using the method [START_REF] Dairay | Numerical dissipation vs. subgrid-scale modelling for large eddy simulation[END_REF][START_REF] Lamballais | Vassilicos: Implicit/explicit spectral viscosity and large-scale SGS effects[END_REF], the validation framework was based on the Taylor-Green Vortex problem as a prototype of wall-free turbulent flow.

As in [START_REF] Dairay | Mesh node distribution in terms of wall distance for large-eddy simulation of wall-bounded flows[END_REF], a regular Cartesian mesh is combined with the use of an immersed boundary technique to discretize the pipe geometry while considering periodic boundary conditions in the three spatial directions. A one-dimensional reconstruction method [START_REF] Gautier | A DNS study of jet control with microjets using an immersed boundary method[END_REF] is used inside the solid to ensure the smoothness of the solution everywhere in the computational domain. This feature clearly improves the accuracy of compact finite difference schemes as used in the code Incompact3d.

Filtered quasi-DNS Database

Quasi-DNS of pipe flow is performed at Re τ = 1000 in order to, afterwards, produce the filtered benchmark database for comparison with ILES results. A mesh [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF] quasi-DNS <u' z u' z > + , DNS [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF] quasi-DNS <u' z u' r > + , DNS [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF] quasi-DNS Fig. 1 Mean velocity (left) and Reynolds stresses profiles (right) for the quasi-DNS at Re τ = 1000 compared to DNS statistics of [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF].

R ij (R-r) + <u' r u' r > + , DNS [3] quasi-DNS <u' θ u' θ > + , DNS
of n x × n y × n z = 768 × 768 × 1920 cells was used, corresponding, in wall units to ∆ x + = ∆ y + = 2.9 and ∆ z + = 13. The term quasi is naturally used due to the employed mesh resolution, however, as shown in figure 1, thanks to the success of the technique, a very good agreement with DNS data of [START_REF] Khoury | Direct numerical simulation of turbulent pipe flow at moderately high reynolds numbers[END_REF] is found.

As in [START_REF] Lamballais | Vassilicos: Implicit/explicit spectral viscosity and large-scale SGS effects[END_REF], the filter is chosen to take the implicit SGS dissipation into account using a spectral Pao-like closure. By specifying the ratio between DNS and LES mesh resolutions, this Pao-like closure provides the spectral transfer functions that are used to define the targeted LES solution. Thus, this targeted solution is determined rigorously and consistently with the artificial dissipation implicitly introduced by the error differentiation in the present ILES framework [START_REF] Dairay | Numerical dissipation vs. subgrid-scale modelling for large eddy simulation[END_REF]. The use of periodic boundary conditions enable the application of the present homogeneous filter defined in the Fourier space. The filter is applied once in every direction to take the anisotropy of the LES mesh into account. Following this procedure, a consistent framework is established for rigorous comparisons between LES and filtered DNS data.

The filter effect over streamwise-z and normal-xy directions separately is shown in figure 2. Acting as a pre-analysis, this information guided the choice of interesting mesh resolutions based on the levels of energy k = 1 2 u i u i to be potentially captured when the viscous and even near-wall production regions are bypassed. It is important to emphasize that the application of filter in normal-xy directions is only possible thanks to the solution smoothness provided by the reconstruction. As shown in figure 3, if no extension of the fluid solution is made into the solid zone before filtering, the no-slip condition is poorly ensured and near-wall region results are not reliable, a problem that is strongly reduced with the reconstruction. Nonetheless, reconstruction's ability to ensure the success of normal direction filtering is limited to not so high values of the filter-width, as shown in figure 2-left by the abnormal near-wall behaviour when data is filtered at a corresponding cell size ∆ x + = ∆ y + = 24. For ∆ x + = ∆ y + = 15, the unrealistic near-wall rising of the turbulent kinetic energy k can be avoided, but persistent velocity fluctuations are observed close to the wall with for instance k + ≈ 0.156 at (Rr) + ≈ 1.03 to be compared to k + ≈ 0.124 for the reference non-filtered value. Here, we have chosen to limit our analysis to ∆ x + = ∆ y + = 12 for which the filtering procedure produces also k + ≈ 0.124 at this near-wall location.

LES Results

Before proceeding to the results, it is important to highlight that the term "implicit wall-layer modelling" is used to emphasize the ability of the present methodology to regularize the solution in the near-wall region without any explicit wall model. The improvement brought by the technique is the simple result of a targeted choice of coefficients for the centred compact finite difference scheme used for the discretization of the viscous term in the Navier-Stokes equations [START_REF] Dairay | Mesh node distribution in terms of wall distance for large-eddy simulation of wall-bounded flows[END_REF].

For a Reynolds number of Re τ = 1000, two implicit LES have been performed using: i) Marginal Resolution: n x × n y × n z = 512 × 512 × 1280 cells, corresponding in wall units to ∆ x + = ∆ y + ≈ 4.5, ∆ z + ≈ 19.5; and ii) Low Resolution: n x × n y × n z = 256 × 256 × 960 cells, expressed in wall units: ∆ x + = ∆ y + = 12, ∆ z + = 26. Mesh resolutions were chosen in order to evaluate the method performance in terms of viscous and near-wall production region bypass respectively. The implicit wall modelling feature of the present approach is evidenced by comparing ILES to no-model LES results performed without any (implicit or explicit) SGS/wall-layer modelling. Beginning with the Marginal Resolution comparisons, in figure 4-right, the deviation between ILES and filtered kinetic energy profiles at the peak points to the existence of numerical/modelling errors. However, the overall precision and agreement of the statistics are the confirmation at high Reynolds number that accurate turbulent statistics can be obtained despite the use of a mesh resolution unable to capture the viscous sublayer. In particular, it can be observed that the peak of k + is remarkably well predicted at a level of detail clearly higher than the minimal scale 4∆ x computed with accuracy. When compared to "No-Model" results, numerical dissipation is found definitely necessary so that it can be seen as playing the role, in this context, of an implicit viscous sublayer modelling. To explain this improvement, it must be mentioned that the no-model solution is subjected to small-scale oscillations which make inaccurate the prediction of friction velocity with an overestimation of about +6%. Here, the use of numerical dissipation can remove these spurious oscillations making clearly more accurate the friction velocity prediction with only a deviation of -1% by comparison to DNS.

For the Low Resolution results, for which the mesh choice was based on the results of figure 2, a good prediction of the mean velocity is recovered by the ILES with an acceptable -2% underprediction of u τ (see figure 5-left). The filtered profile of k + in figure 5-right shows that good level of energy could still be captured despite the astonishing value of 12 wall units used for the normal-xy discretization. Even though the profile of k + is somewhat irregular in the region of the peak, its amplitude is only underestimated by -11% with a shape captured at a scale evidently smaller than the minimum scale of accuracy 4∆ x + = 48. In the outer layer, the agreement with filtered DNS is almost perfect. The same cannot be said about the no-model LES for which both mean velocity and kinetic energy profiles are completely unrealistic when such a coarse mesh resolution is used. Based on these observations, it can be concluded that the present implicit SGS modelling has also a feature of implicit wall-layer model through its ability to restore the quality of the wall friction from +24% to -2% while enabling realistic prediction of the fluctuating velocity profiles. Note however that for the present Low Resolution, near-wall turbulent statistics are subjected to errors related to the azimuthal loca- tion of the nodes, especially those close to the cardinal points. These errors, due to the immersed boundary technique, tend to compensate each-other leading to the acceptable k + profile presented in figure 5-right. A development is under progress to remove these immersed boundary artefacts making a room for further improvement of the present implicit wall-layer modelling.

Fig. 2

 2 Fig. 2 Turbulent kinetic energy from filtered quasi-DNS at different mesh resolutions. The drop of energy is analysed for filtering along (left) streamwise-z direction and (right) normal-xy directions.

Fig. 3

 3 Fig.3Comparison of mean (left) and instantaneous (right) velocity profiles demonstrates the effect/necessity of reconstruction on the reliability of data filtered in the normal Cartesian directions (x, y). In this illustration, quasi-DNS data have been filtered at a corresponding mesh resolution of ∆ x + = ∆ y + = 12.

Fig. 4

 4 Fig. 4 Mean velocity profile (left) and Reynolds stresses profiles (right) for the Marginal Resolution ILES at Re τ = 1000. Results are compared to: no-model LES, unfiltered and filtered quasi-DNS statistics.

Fig. 5

 5 Fig. 5 Mean velocity profile (left) and Reynolds stresses profiles (right) for the low resolution ILES at Re τ = 1000. Results are compared to: no-model LES, unfiltered and filtered quasi-DNS statistics.

In the discrete and inviscid sense up to the time advancement error.

Acknowledgements This work was granted access to the HPC resources of TGCC/CINES/IDRIS under the allocation A0012A07624/A0032A07624/A0052A07624 made by GENCI.