Eric Djeutcha

Jules Sadefo Kamdem

Louis Aimé Fono

Mixed Modified Fractional Merton model of the bear spread Basket put option using the multidimensional Mellin transform

In this paper, The generalized Mixed-Modified-Fractional-Merton like partial differential equation with multi-assets under mixed modified fractional geometric Brownian motion was derived. The multidimensional Mellin transform was applied to derive the integral equation for the price of the European put option on a bear spread basket of multi-assets.

Introduction

Mandelbrot et al. [START_REF] Mandelbrot | Fractional Brownian motions, fractional noises and applications[END_REF] was introduced the Fractional Brownian Motion (FBM) as an extension of the Brownien motion. In recent decades, several researchers including, Willinger and Taqqu [START_REF] Willinger | Stock market prices and long-range dependence[END_REF], Lin [START_REF] Lin | Stochastic analysis of fractional Brownian motions[END_REF] et Rosteck [START_REF] Rostek | Option Pricing in Fractional Brownian Markets[END_REF] have focussed their studies on this process by studying self-similarity and the property of long-range dependency. Those authors use FBM to express the random part of financial models. However, the existence of this process in financial models can lead to some difficulties. We can show that except the case H = 1 2 , the FBM process is neither a Markov process, nor a semimartingale. Therefore, Roger [START_REF] Rogers | Arbitrage with fractional Brownian motion[END_REF] and Cheridito [START_REF] Cheridito | Arbitrage in fractional Brownian motion models[END_REF] have shown that there can be no equivalent probability under which the process becomes a local martingale, this raises the arbitrage problem. To overcome this problem, Cheridito [START_REF] Cheridito | Mixed fractional Brownian motion[END_REF] has applied the Mixed Fractional Brownian Motion (MFBM) and Zlli [START_REF] Zili | On the mixed fractional brownian motion[END_REF] has applied its generalisation to unveil the random part of the financial model. A MFBM process whose parameters a, b and H is a linear combination of Brownian Motion B t and independent Fractional Brownian Motion B H t , defined on a probability space (Ω, F, P) by:

M H,a,b t = M H t = aB t + bB H t , ∀(a, b) ∈ R * + × R + , t ∈ [0, T ]. (1)
Where H ∈ (0, 1) is the Hurst parameter. Cheridito [START_REF] Cheridito | Mixed fractional Brownian motion[END_REF] showed that the MFBM process with H ∈] 3 4 , 1[ equals to a martingale B t and is therefore arbitrage-free. Nonetheless, there is no equivalent probability under which the MFBM for all H ∈] 1 2 , 3 4 [ becomes a local martingale. Djeutcha et al. [START_REF] Djeutcha | Solving Arbitrage Problem on the Financial Market Under the Mixed Fractional Brownian Motion With Hurst Parameter H ∈]1/2[END_REF] suggest a new process defined on the same probability space (Ω, F, P) by:

M H,ε t = aB t + bB H,ε t , ∀(a, b) ∈ R * + × R + , H ∈] 1 2 , 1[, (2)
with

B H,ε t = t 0 (t -s + ε) H-1 2 dBs, ∀ε ∈ R * + , t ∈ [0, T ]. (3)
Where B s is a Brownian Motion and ε is a positive constant. The process (2) is an modification of process (1), framed MMFBM stands for "Mixed Modified Fractional Brownian Motion", its importance is linked to fact that it is a continous martingale. It is used in this paper to design the financial assets.The Black Scholes Model [START_REF] Black | The pricing of option and corporate liabilities[END_REF], assumes the price of heavily traded assets follows a geometric Brownian motion with constant drift and volatility. It model cannot capture many of the features of asset price returns in the absence of dividend payments, Merton [START_REF] Merton | Option pricing when underlying stock returns are discontinuous[END_REF] considered a jump-diffusion process that allows for the probability of the asset price to change at large magnitudes irrespective of the time interval between successive observations. Empirical studies have showed that the asset price is better described by a process with a discontinuous sample path (Rosenfeld [16], Jarrow and Rosenfeld [START_REF] Jarrow | Jump risks and the intertemporal capital asset pricing model[END_REF], Ball and Torous [START_REF] Ball | On jumps in common stock prices and their impact on call option pricing[END_REF], and Brown and Dybvig [START_REF] Brown | The empirical implications of the Cox-Ingersoll-Ross theory of the term structure of interest rates[END_REF]). Cont and Tankov [START_REF] Cont | Financial Modelling with Jump Processes[END_REF] showed that unlike standard diffusion models such as Black Scholes [START_REF] Black | The pricing of option and corporate liabilities[END_REF], jump-diffusion models produce rich structures of the distribution of asset returns and implied volatility surfaces. To account for the possibility of instantaneous jumps in the asset price, Merton [START_REF] Merton | Option pricing when underlying stock returns are discontinuous[END_REF] proposed the following modification of Black Scholes [START_REF] Black | The pricing of option and corporate liabilities[END_REF] by assuming that the discontinuous jumps arrive as a Poisson process.

dS(t) = S(t)[r -λE(Y -1))]dt + σS(t)dW (t) + σS(t)(Y -1)dN (t) (4)
where Y (t) is a nonnegative continuous random variable with Y -1 denoting the impulse change in the asset price from S(t) to Y S(t) as a result of the jump and Y -1 ∼ N (µ, δ 2 ), E is the expectation operator, and N = (N (t), t 0) is a Poisson process with constant intensity λ and such that dN (t) = 1 (respectively, dN (t) = 0) with probability λdt (respectively, 1 -λdt). Using Mellin transform techniques adapted from Rodrigo and Mamon [START_REF] Rodrigo | An application of Mellin transform techniques to a Black-Scholes equation problem[END_REF], Li and Rodrigo [START_REF] Li | Alternative results for option pricing and implied volatility in jump-diffusion models using Mellin transforms[END_REF] studied the Partial Integro Differantial Equation (PIDE) and found exact pricing formulas for European options with general payoffs such as those in many types of option. A PIDE analogue of Dupire's PDE was also derived and used to find an explicit formula for the implied volatility. In the same vein, Rodrigo and Goard [START_REF] Rodrigo | Pricing of general European options on discrete dividend-paying assets with jump-diffusion dynamics[END_REF] considered PIDE with a time varying D and obtained exact pricing formulas for European options on discrete dividend-paying assets.

An option is a financial derivative that grants its holder the right without obligation to buy a specific asset on or before given date in the future for an agreed price called the exercise price. A multi-asset option is path dependent option that depends on more than one underlying asset with the payoff defined by a function of the asset prices. A basket option is a financial contract with a portfolio of several underlying equity assets.The payoff function for a basket put option is of the form:

(K - n i=1 α i S i ) + (5)
where α i is the number of shares of asset i in the basket. S i is the price of asset i in the basket, K is the exercise price and

n i=1 α i = 1.
Several authors have studied the Basket option, among which Xu and Zheng [START_REF] Xu | Basket options valuation for a local volatility jump-diffusion model with asymptotic expansion method[END_REF] who have studied the Basket options valuation for a local volatility jump-diffusion model with asymptotic expansion method, Kamdem [START_REF] Kamdem | Option pricing with Levy process using Mellin Transform[END_REF], use Mellin transform to get the expression for the free boundary an price of an American finite-lived option, when the underlying is govern by the Levy process and show how to compute the price of an American option on a basket of stocks using Mellin transform of several variables. A bear spread option is a vertical spread. It consists of buying one put in hopes of profiting from a decline in the underlying stock, and writing another put with the same expiration but with a lower strike price, as a way to offset some of the cost. The payoff function of the bear spread option is given by

(K 2 -x) + -(K 1 -x) + (6)
Using Eq.( 5) and Eq.( 6), we define the Bear spread Basket put option as follows

(K 2 - n i=1 α i S i ) + -(K 1 - n i=1 α i S i ) + (7)
This paper add a long-range dependency property to the model given by Merton [START_REF] Merton | Option pricing when underlying stock returns are discontinuous[END_REF] as far as the non-arbitrage condition is concerned.

Our approach will consist firstly of describing the random part of the Merton model by using the Mixed Modified Fractional Brownian Motion and secondly to evaluate the basket option coupled with the bear spread option using the Mellin transform method with several variables and drawing inspiration from research of Kamdem [START_REF] Kamdem | Option pricing with Levy process using Mellin Transform[END_REF], Frontczak [START_REF] Frontczak | Pricing Options in Jump Diffusion Models Using Mellin Transforms[END_REF] and Fadugba [START_REF] Fadugba | Mellin transform in higher dimensions for the valuation of the European basket put option with multi-dividend paying stocks[END_REF].

Our motivation comes from the fact that the study of the markets with the MMFBM process is recent and the evaluation of the bear spread basket put option price whose the payoff function is given by Eq.( 7) seems to be not studied in the recent litterature.

The main concern in this paper will be centered around finding the partial differential equation of the pricing Basket options under MMFM model (Mixed Modified Fractional Merton model). We also determine the closed formula for the bear spread Basket put option price. The section 2 of this paper presents the Mixed Modified Fractional Merton Model, the section 3 defined the Bear spread Basket put options in a MMFM model, the section 4 establish the Partial Integro Differential Equation (PIDE) under the MMFM model and the section 5 gives the Multi-dimensional Mellin transform for the tarification of the european Put option on a bear spread basket of multi-assets.

Presentation of the MMFM model

We recall that the Merton model is defined by Eq.( 4). However, to resolve the arbitrage problem in such a market when H ∈] 1 2 , 1[, we plan to substitue in (4) the process W t by the MMFBM defined by equation [START_REF] Ball | On jumps in common stock prices and their impact on call option pricing[END_REF]. The latter is thus called "Mixed Modified Fractional Merton model" in short MMFM model and is a modification of model (4) defined by Eq.( 8) below.

Definition 1. The MMFM model is one in which the dynamic underlying asset price S ε t verify the following stochastic equation:

dS ε t S ε t = [r -λE(Y -1)]dt + σdM H,ε t + σ(Y -1)dN (t), S ε 0 = S 0 > 0, H ∈] 1 2 , 1[ (8)
By using the equations (2)-( 3) and taking into consideration that :

dM H,ε t = bφ ε t dt + a + bε H-1 2 dB t (9)
with

φ ε t = H - 1 2 t 0 (t -s + ε) H-3 2 dB s , (10)
Eq.( 8) can also be written as

dS ε t S ε t = [r -λE(Y -1) + bσφ ε t ]dt + σ(a + bε H-1 2 )dW t + σ(Y -1)dN (t). (11)
In the remainder of this paper, the MMFM Model given by Eq.( 8) is used for the tarification of the bear spread Basket options. The solution of Eq.( 8) is

S ε t = S ε 0    r -λλE(Y -1) - σ 2 2 a + bε H-1 2 2 t + σM H,ε t + N (t) i=1 Y i    (12)
3 Bear spread Basket options in a MMFM model

In this section, we now shall concentrate on the pricing of bear spread basket options in the MMFM model. Denoting by S ε i,t the price of the i-th asset in the basket at time t, the basket is given by

S ε t = n i=1 α j S ε i,t , (13)
where α j , for all j is deterministic, positive and constant weights specified by the option contract. We assume that the risk neutral measure

dS ε i,t S ε i,t = (r -λE(Y -1))dt + σ i dM H,ε i,t + (Y -1)dN (t), (14)
where (M H,ε i,t ) t 0 is a Mixed Modified Fractional Brownian motion associated with the price process of asset i. Given the above dynamics, the i-th asset price at time t equals

S ε i,t = S ε i,0    r -λλE(Y -1) - σ 2 i 2 a + bε H-1 2 2 t + σ i M H,ε i,t + N (t) i=1 N (t i )    (15)
Further, we assume the different asset prices to be instantaneously correlated according to

corr(dM H,ε i,t , dM H,ε j,t ) = ρ ij (a + bε H-1 2 ) 2 dt, ∀i = j, (16)
with ρ ∈ [-1, 1]. Using Eq.( 17) the Mixed Modified fractional Bear spread Basket put option is defined as follows

φ(S ε 1 , S ε 2 , • • • , S ε n ) = (K 2 - n i=1 α i S ε i ) + -(K 1 - n i=1 α i S ε i ) + (17)
where K 1 and K 2 are respectively the strike price of the option such that K 1 < K 2 .

PDE of the option price under the MMFM model

We consider an option on n stocks, with each the stock price S ε i , i = 1, 2, 3, • • • , n follow correlated geometric Mixed Modified Fractional Brownian motion with drift µ i and volatility σ i solution of Eq.( 14). Before the derivation of the generalized Mixed-Modified-Fractional-Merton partial differential equation, a counterpart; the Itö's lemma in higher dimensions is needed. Consider a multi-dimensional Mixed Modified Fractional Itö's process of the form

dX i,t = µ i (X i,t , t)dt + σ i (X i,t , t)dM H,ε i,t , 1 i n, (18)
Therefore, the multi-dimensional Itö's lemma for Eq.( 18) is given by

df (X t , t) =   ∂f (X t , t) ∂t + n i=1 µ i ∂f (X t , t) ∂x i + 1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j x j x j ∂ 2 f (X t , t) ∂x i ∂x j   dt+ n i=1 σ i ∂f (X t , t) ∂x i dM H,ε i,t (19)
where f (X t , t) ∈ C 2,1 . The derivation of the generalized Mixed-Modified-Fractional-Merton partial differential equation for the European bear spread basket put option is given in the following result.

Theorem 1. Let f (S ε 1 , S ε 2 , S ε 3 , • • • , S ε n , t) ∈ C 2,1
be some functions with multi-assets, S ε i be the current price of the underlying asset i , S ε j be the current price of the underlying asset j , K 1 and K 2 be the strike price of the bear spread option, σ i be the volatility of asset i , σ j be the volatility of asset j , ρ ij be the correlation coefficient between asset i and asset j , T be the time to expiry, r be the risk-free interest rate and . Using the Itö's lemma [START_REF] Rostek | Option Pricing in Fractional Brownian Markets[END_REF],the generalized Mixed-Modified-Fractional-Merton partial differential equation for price of an European bear spread basket put option is obtained as

∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + 1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j + λE [f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t)] + n i=1 (r -λE(Y -1))S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i -rf (S ε 1 , S ε 2 , • • • , S ε n , t) = 0 (20)
Proof. Construct a portfolio consisting of an option

f (S ε 1 , S ε 2 , • • • , S ε n , t) and ∂f (S ε 1 ,S ε 2 ,••• ,S ε n ,t) ∂S ε i amount of assets S ε i .
Let the asset j S be driven by correlated geometric Brownian motions. Using Eq.( 19) for some functions

f (S ε 1 , S ε 2 , • • • , S ε n , t) with multi-assets,yields df (X t , t) = ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + n i=1 µ i S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dt +   1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j   dt + n i=1 σ i S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dM H,ε i,t + [f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t)] dN (t) (21)
The value of the portfolio denoted by Π is given by

Π = f (S ε 1 , S ε 2 , • • • , S ε n , t) - n i=1 S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (22)
After one time step dt, the value of the portfolio changes by

dΠ = df (S ε 1 , S ε 2 , • • • , S ε n , t) - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dS ε i (23)
Substituting Eq.( 14) and Eq.(25) into Eq.( 23) yields

dΠ = df (S ε 1 , S ε 2 , • • • , S ε n , t) - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dS ε i (24) dΠ = ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + n i=1 (r -λE(Y -1))S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dt +   1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j   dt + n i=1 σ i S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dM H,ε i,t - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (r -λE(Y -1))S ε i dt - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i σ i S ε i dM H,ε i,t - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (Y -1)S ε i dN (t) + [f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t)] dN (t)) (25)
i.e

dΠ =   ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + 1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j   dt + f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t) - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (Y -1)S ε i dN (t)) (26)
or dN (t) = 1 with the probability λdt. Therefore, we have

dΠ =   ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + 1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j   dt + λE f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t) - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (Y -1)S ε i dt (27)
The portfolio is now riskless due to the elimination of dM Hε t term. It must then earn a return similar to other short term riskless securities such as bank account. Therefore

dΠ = rΠdt ( 28
)
where r is the riskless interest rate. Substituting Eq.( 22) and Eq.( 27) into Eq.(28) gives (56)

  ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂t + 1 2 (a + bε H-1 2 ) 2 n j,i=1 ρ ij σ i σ j S ε i S ε j ∂ 2 f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂x i ∂x j   dt + λE f (S ε 1 y, • • • , S ε n y, t) -f (S ε 1 , • • • , S ε n , t) - n i=1 ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i (Y -1)S ε i dt = r f (S ε 1 , S ε 2 , • • • , S ε n , t) - n i=1 S ε i ∂f (S ε 1 , S ε 2 , • • • , S ε n , t) ∂S ε i dt (29) with                                              d 1 = ln    n i=1 α i S ε i,t K 2   +
The proof of this Proposition 1 is obvious and as indication, we can use an approach given by Frontczak [START_REF] Frontczak | Pricing Options in Jump Diffusion Models Using Mellin Transforms[END_REF].

Conclusion

In this paper, we have studied the generalized Mixed-Modified-Fractional-Merton like partial differential equation with multi-assets under mixed modified fractional geometric Brownian motion was derived. We have used the multidimensional Mellin transform was applied to derive the integral equation for the price of the European put option on a bear spread basket of multi-assets.

  r(T -t)-λ(Y -1)(T -t)+n ln(Y )+ 1 2 (a+bε H-1 2 ) 2

					n i=1 σ 2 i (T -t)+nδ 2
				(a+bε H-1 2 ) 2	n i=1 σ 2 i (T -t)+nδ 2
	d 2 = d 1 -(a + bε H-1 2 ) 2	n i=1 σ 2 i (T -t) + nδ 2
	ln   	n i=1 α i S ε i,t K 1	  + r(T -t)-λ(Y -1)(T -t)+n ln(Y )+ 1 2 (a+bε H-1 2 ) 2	n i=1 σ 2 i (T -t)+nδ 2
	d 3 =			(a+bε H-1 2 ) 2	n i=1 σ 2 i (T -t)+nδ 2
	d 4 = d 3 -(a + bε H-1 2 ) 2	n i=1 σ 2 i (T -t) + nδ 2 .

Rearranging and solving the last equation further, the multi-dimensional Mixed Modified Fractional Merton partial differential equation for the price of an option f (S ε 1 , S ε 2 , • • • , S ε n , t) with multi-assets is obtained as

5 Multi-dimensional Mellin transform for the tarification of the european Put option on a bear spread basket of multi-assets

The following result gives the derivation of the expression for the integral equation for the price of the European bear spread basket put option by means of the Mellin transform in higher dimensions.

Theorem 2. The generalized Mixed-Modified-Fractional-Merton partial differential equation for the price of the European bear spread basket put option is given by

with the initial and boundary conditions are given by

Then, the expression for the integral equation for the price of the European put option on a bear spread basket of multi-assets is obtained as

Proof. It is assumed that the underlying assets and follow geometric Mixed Modified Fractional Brownian motion. From Eq.( 31)

To use the multi-dimensional Mellin transform, assume that f

Equation Eq.(37) guarantees the existence of the multi-dimensional Mellin transform. Let

where the complex variable w 1 , w 2 , • • • , w n exists in an appropriate domain of convergence C n . Conversely, the multi-dimensional Mellin transform inversion formula for the European put option on a bear spread basket of multi-assets is given by

Thus to find the multi-dimensional Mellin transform of Eq.( 31), applying the definition given by Eq.(38) to Eq.( 31), yields

where

We recall that, for the jump size distribution, different distributions are commonly used in financial literature, pominent candidates are among others(Log-normally distributed jumps,double exponentially distributed jumps, Gamma distributed jumps and One-sided exponentially distributed jumps). The choice of the distribution of Y is crucial in order to solve the problem analytically in the (S ε 1 , S ε 2 , • • • , S ε n , t)-domain. In this paper, the explicit expression for G(w 1 , w 2 , • • • , w n ) is given in the case for Y follows the Log-normally distributed jumps given by Merton [START_REF] Merton | Option pricing when underlying stock returns are discontinuous[END_REF]. We obtain :

Therefore

By setting

The solution to Eq.( 45) is obtained as

where J(w 1 , w 2 , • • • , w n ) is the constant of integration given by

The multi-dimensional Mellin transform of the final condition

in Eq.( 33) is obtained as

where

Equation Eq.( 50) is called the multinomial beta function of n-variables. Substituting Eq.(49) into Eq.( 47) yields

Therefore, Eq.( 46) becomes

Taking the inverse multi-dimensional Mellin transform of Eq.( 52), the expression for the integral equation for the price of the European put option on a bear spread basket of multi-assets is obtained as

We complete this section by showing explicitly that Eq. (39) with the special choices of f

) and log-normally distributed jumps is equivalent to Mixed Modified Fractional Merton's infinite series solution for a European put in the bear spread basket option.

and jumps are distributed log-normally with parameters µ and δ, Y ∼ ln(µ, δ), then Eq.( 39) is equivalent to Mixed Modified Fractional Merton's solution which is formulated as an infinite series of Mixed Modified Fractional Black Schole Merton-prices(MMFBSM):

Where f M M F BSM (S ε 1 , • • • , S ε n , K 1 , K 2 , r, T, t) denotes the European put in the bear spread basket option due to the Mixed Modified Fractional Black Schole Merton and equals


F Ben Saïd

J.-L Nicolas
email: jlnicola@in2p3.fr

A Zekraoui
email: ahlemzekraoui@yahoo.fr

On the parity of generalized partition functions III by

Keywords: Partitions, periodic sequences, order of a polynomial, orbits, 2-adic numbers, counting function, Selberg-Delange formula

, the elements of the set A = A(1 + z + z 3 + z 4 + z 5 ), for which the partition function p(A, n) (i.e. the number of partitions of n with parts in A) is even for all n ≥ 6 are determined. An asymptotic estimate to the counting function of this set is also given.

1 Introduction.

Let N (resp. N 0 ) be the set of positive (resp. non-negative) integers. If A = {a 1 , a 2 , ...} is a subset of N and n ∈ N then p(A, n) is the number of partitions of n with parts in A, i.e., the number of solutions of the diophantine equation

a 1 x 1 + a 2 x 2 + . . . = n, (1.1)
in non-negative integers x 1 , x 2 , .... As usual we set p(A, 0) = 1.

The counting function of the set A will be denoted by A(x), i.e.,

A(x) =| {n ≤ x, n ∈ A} | . (1.2)
Let F 2 be the field with 2 elements, P = 1 + ǫ 1 z 1 + ...

+ ǫ N z N ∈ F 2 [z], N ≥ 1.
Although it is not difficult to prove (cf. [START_REF] Nicolas | On the parity of additive representation functions[END_REF], [START_REF] Saïd | On some sets with even partition function[END_REF]) that there is a unique subset 1 A = A(P ) of N such that the generating function F (z) satisfies

F (z) = F A (z) = a∈A 1 1 -z a = n≥0
p(A, n)z n ≡ P (z) (mod 2), (1.3) the determination of the elements of such sets for general P ′ s seems to be hard.

Let the decomposition of P into irreducible factors over F 2 be P = P α 1 1 P α 2 2 ...P α l l .

(1.4)

We denote by β i = ord(P i ), 1 ≤ i ≤ l, the order of P i , that is the smallest positive integer β i such that P i (z) divides 1 + z β i in F 2 [z]. It is known that β i is odd (cf. [START_REF] Lidl | Introduction to finite fields and their applications[END_REF]). We set β = lcm(β 1 , β 2 , ..., β l ).

(1.5)

Let A = A(P ) satisfy (1.3) and σ(A, n) be the sum of the divisors of n belonging to A, i.e.,

σ(A, n) = d|n, d∈A d = d|n dχ(A, d), (1.6)
where χ(A, .) is the characteristic function of the set A, i.e, χ(A, d) = 1 if d ∈ A and χ(A, d) = 0 if d ∈ A. It was proved in [START_REF] Saïd | Sets of parts such that the partition function is even[END_REF] (see also [START_REF] Saïd | On a conjecture of Nicolas-Sárközy about partitions[END_REF], [START_REF] Lahouar | Fonctions de partitions à parité périodique[END_REF]) that for all k ≥ 0, the sequence (σ(A, 2 k n) mod 2 k+1 ) n≥1 is periodic with period β defined by (1.5), in other words,

n 1 ≡ n 2 (mod β) ⇒ ∀k ≥ 0, σ(A, 2 k n 1 ) ≡ σ(A, 2 k n 2 ) (mod 2 k+1 ). (1.7)
Moreover, the proof of (1.7) in [START_REF] Saïd | Sets of parts such that the partition function is even[END_REF] allows to calculate σ(A, 2 k n) mod 2 k+1 and to deduce the value of χ(A, n) where n is any positive integer. Indeed, let S A (m, k) = χ(A, m) + 2χ(A, 2m) + . . . + 2 k χ(A, 2 k m).

(1.8)

If n writes n = 2 k m with k ≥ 0 and m odd, (1.6) implies In the above sums, n d is always a multiple of 2 k , so that, from the values of σ(A, n d ), by (1.10), one can determine the value of S A (m, k) mod 2 k+1 and by (1.8), the value of χ(A, 2 i m) for all i, i ≤ k.

σ(A, n) = σ(A, 2 k m) = d | m dS A (d, k), (1.9
Let β be an odd integer ≥ 3 and (Z/βZ) * be the group of invertible elements modulo β. We denote by < 2 > the subgroup of (Z/βZ) * generated by 2 and consider its action ⋆ on the set Z/βZ given by a ⋆ x = ax for all a ∈ < 2 > and x ∈ Z/βZ. The quotient set will be denoted by (Z/βZ)/ <2> and the orbit of some n in Z/βZ by O(n). For P ∈ F 2 [z] with P (0) = 1 and ord(P ) = β, let A = A(P ) be the set obtained from (1.3). Property (1.7) shows (after [START_REF] Baccar | On the divisor function of sets with even partition functions[END_REF]) that if n 1 and n 2 are in the same orbit then σ(A, 2 k n 1 ) ≡ σ(A, 2 k n 2 ) (mod 2 k+1 ), ∀k ≥ 0.

(1.11)

Consequently, for fixed k, the number of distinct values that (σ(A, 2 k n) mod 2 k+1 ) n≥1 can take is at most equal to the number of orbits of Z/βZ.

Let ϕ be the Euler function and s be the order of 2 modulo β, i.e., the smallest positive integer s such that 2 s ≡ 1 (mod β). If β = p is a prime number then (Z/pZ) * is cyclic and the number of orbits of Z/pZ is equal to 1 + r with r = ϕ(p) s = p-1 s . In this case, we have (Z/pZ)/ <2> = {O(g), O(g 2 ), ..., O(g r ) = O(1), O(p)}, (1.12) where g is some generator of (Z/pZ) * . For r = 2, the sets A = A(P ) were completely determined by N. Baccar, F. Ben Saïd and J.-L. Nicolas ( [START_REF] Baccar | On sets such that the partition function is even from a certain point on[END_REF], [START_REF] Saïd | Even partition functions[END_REF]). Moreover, N. Baccar proved in [START_REF] Baccar | Sets with even partition functions and 2-adic integers[END_REF] that for all r ≥ 2, the elements of A of the form 2 k m, k ≥ 0 and m odd, are determined by the 2-adic development of some root of a polynomial with integer coefficients. Unfortunately, his results are not explicit and do not lead to any evaluation of the counting function of the set A. When r = 6, J.-L. Nicolas determined (cf. [START_REF] Nicolas | On the parity of generalized partition functions II[END_REF]) the odd elements of A = A(1 + z + z 3 + z 4 + z 5 ). His results ( which will be stated in Section 2, Theorem 0) allowed to deduce a lower bound for the counting function of A. In this paper, we will consider the case p = 31 which satisfies r = 6. In F 2 [z], we have 1 -z 31 1 -z = P (1) P (2) ...P (6) , (1.13) with P (1) = 1+z+z 3 +z 4 +z 5 , P (2) = 1+z+z 2 +z 4 +z 5 , P (3) = 1+z 2 +z 3 +z 4 +z 5 , P (4) = 1 + z + z 2 + z 3 + z 5 , P (5) = 1 + z 2 + z 5 , P (6) = 1 + z 3 + z 5 .

In fact, there are other primes p with r = 6. For instance, p = 223 and p = 433.

In Section 2, for A = A(P (1) ), we evaluate the sum S A (m, k) which will lead to results of Section 3 determining the elements of the set A. Section 4 will be devoted to the determination of an asymptotic estimate to the counting function A(x) of A. Although, in this paper, the computations are only carried out for P = P (1) , the results could probably be extended to any P (i) , 1 ≤ i ≤ 6, and more generally, to any polynomial P of order p and such that r = 6.

Notation. We write a mod b for the remainder of the euclidean division of a by b. The ceiling of the real number x is denoted by

⌈x⌉ = inf{n ∈ Z, x ≤ n}. . 2 The sum S A (m, k), A = A(1 + z + z 3 + z 4 + z 5 ).
From now on, we take A = A(P ) with

P = P (1) = 1 + z + z 3 + z 4 + z 5 .
(

The order of P is β = 31. The smallest primitive root modulo 31 is 3 that we shall use as a generator of (Z/31Z) * . The order of 2 modulo 31 is s = 5 so that

(Z/31Z)/ <2> = {O(3), O(3 2 ), ..., O(3 6 ) = O(1), O(31)}, (2.2)
with

O(3 j ) = {2 k 3 j , 0 ≤ k ≤ 4}, 1 ≤ j ≤ 6 (2.3) and O(31) = {31n, n ∈ N}. (2.4)
For k ≥ 0 and 0 ≤ j ≤ 5, we define the integers u k,j by

u k,j = σ(A, 2 k 3 j ) mod 2 k+1 . (2.5)
The Graeffe transformation. Let K be a field and K[[z]] be the ring of formal power series with coefficients in K. For an element

f (z) = a 0 + a 1 z + a 2 z 2 + . . . + a n z n + . . . of this ring, the product f (z)f (-z) = b 0 + b 1 z 2 + b 2 z 4 + . . . + b n z 2n + . . .
is an even power series. We shall call G(f ) the series

G(f )(z) = b 0 + b 1 z + b 2 z 2 + . . . + b n z n + . . . . (2.6)
It follows immediately from the above definition that for f, g

∈ K[[z]], G(f g) = G(f )G(g). (2.7)
Moreover if q is an odd integer and f (z) = 1 -z q , we have G(f ) = f . We shall use the following notation for the iterates of f by G :

f (0) = f, f (1) = G(f ), . . . , f (k) = G(f (k-1) ) = G (k) (f ). (2.8)
More details about the Graeffe transformation are given in [START_REF] Saïd | Sets of parts such that the partition function is even[END_REF]. By making the logarithmic derivative of formula (1.3), we get (cf. [START_REF] Nicolas | On the parity of additive representation functions[END_REF]) :

∞ n=1 σ(A, n)z n = z F ′ (z) F (z) ≡ z P ′ (z) P (z) (mod 2), (2.9)
which, by Propositions 2 and 3 of [START_REF] Saïd | Sets of parts such that the partition function is even[END_REF], leads to

∞ n=1 σ(A, 2 k n)z n ≡ z P ′ (k) (z) P (k) (z) = z 1 -z 31 P ′ (k) (z)W (k) (z) (mod 2 k+1
), (2.10) with P ′ (k) (z) = d dz (P (k) (z) and W (z) = (1 -z)P (2) (z)...P (6) (z).

(2.11) Formula (2.10) proves (1.11) with β = 31, and the computation of the kth iterates P (k) and W (k) by the Graeffe transformation yields the value of σ(A, 2 k n) mod 2 k+1 . For instance, for k = 11, we obtain :

u k,0 = 1183, u k,1 = 1598, u k,2 = 1554, u k,3 = 845, u k,4 = 264, u k,5 = 701.
A divisor of 2 k 3 j is either a divisor of 2 k-1 3 j or a multiple of 2 k . Therefore, from (2.5) and (1.6), u k,j ≡ u k-1,j (mod 2 k ) holds and the sequence (u k,j ) k≥0 defines a 2-adic integer U j satisfying for all k ′ s :

U j ≡ u k,j (mod 2 k+1 ), 0 ≤ j ≤ 5.
(2.12)

It has been proved in [START_REF] Baccar | Sets with even partition functions and 2-adic integers[END_REF] that the U ′ j s are the roots of the polynomial R(y) = y 6 -y 5 + 3y For convenience, if j ∈ Z, we shall set

U j = U j mod 6 . (2.14)
We define the completely additive function ℓ :

Z \ 31Z → Z/6Z by ℓ(n) = j if n ∈ O(3 j ), (2.15)
so that ℓ(n 1 n 2 ) ≡ ℓ(n 1 ) + ℓ(n 2 ) (mod 6). We split the odd primes different from 31 into six classes according to the value of ℓ. More precisely, for 0 ≤ j ≤ 5,

p ∈ P j ⇐⇒ ℓ(p) = j ⇐⇒ p ≡ 2 k 3 j (mod 31), k = 0, 1, 2, 3, 4.
(2.16)

We take L : N \ 31N -→ N 0 to be the completely additive function defined on primes by L(p) = ℓ(p).

(2.17)

We define, for 0 ≤ j ≤ 5, the additive function ω j : N -→ N 0 by

ω j (n) = p|n, p∈P j 1 = p|n, ℓ(p)=j 1, (2.18)
and ω(n) = ω 0 (n) + ... + ω 5 (n) = p|n 1. We remind that additive functions vanish on 1.

From (2.5), (2.3), (1.11) and (2.12), it follows

that if n = 2 k m ∈ O(3 j ) (so that j = ℓ(n) = ℓ(m)), σ(A, n) = σ(A, 2 k m) ≡ U ℓ(m) (mod 2 k+1 ).
(2.19)

We may consider the 2-adic number

S(m) = S A (m) = χ(A, m) + 2χ(A, 2m) + ... + 2 k χ(A, 2 k m) + ... (2.20) satisfying from (1.8), S(m) ≡ S A (m, k) (mod 2 k+1 ). (2.21) Then (1.10) implies for (m, 31) = 1, mS(m) = d | m µ(d)U ℓ( m d ) . (2.22)
If 31 divides m, it was proved in [3, (3.6)] that, for all k ′ s,

σ(A, 2 k m) ≡ -5 (mod 2 k+1 ). (2.23) Remark 1.
No element of A has a prime factor in P 0 . This general result has been proved in [START_REF] Baccar | On the divisor function of sets with even partition functions[END_REF], but we recall the proof on our example : let us assume that n = 2 k m ∈ A, where m is an odd integer divisible by some prime p in P 0 , in other words ω 0 (m) ≥ 1. (1.10) gives

mS A (m, k) = d | m µ(d)σ A, n d = d | m µ(d)σ A, 2 k m d = d | m p µ(d)σ A, 2 k m d + d | m p µ(pd)σ A, 2 k m pd = d | m p µ(d) σ A, 2 k m d -σ A, 2 k m pd .
In the above sum, both m d and m pd are in the same orbit, so that from (1.11),

σ(A, 2 k m d ) ≡ σ(A, 2 k m pd ) (mod 2 k+1 ) and therefore mS A (m, k) ≡ 0 (mod 2 k+1 ). Since m is odd and (cf. (1.8)) 0 ≤ S A (m, k) < 2 k+1 then S A (m, k) = 0, so that by (1.8), 2 h m ∈ A, for all 0 ≤ h ≤ k.
In [START_REF] Nicolas | On the parity of generalized partition functions II[END_REF], J.-L. Nicolas has described the odd elements of A. In fact, he obtained the following : Theorem 0. ( [START_REF] Nicolas | On the parity of generalized partition functions II[END_REF])) (a) The odd elements of A which are primes or powers of primes are of the form p λ , λ ≥ 1, satisfying one of the following four conditions : p ∈ P 1 and λ ≡ 1, 3, 4, 5 (mod 6) p ∈ P 2 and λ ≡ 0, 1 (mod 3) p ∈ P 4 and λ ≡ 0, 1 (mod 3) p ∈ P 5 and λ ≡ 0, 2, 3, 4 (mod 6). (c) An odd element n ∈ A satisfies ω 0 (n) = 0 and ω 3 (n) = 0 or 1 ; in other words, n is free of prime factor in P 0 and has at most one prime factor in P 3 .

(b) No odd element of A is a multiple of 31 2 . If m is odd, m = 1,
(d) The odd elements of A different from 1, not divisible by 31, which are not primes or powers of primes are exactly the odd n ′ s, n = 1, such that (where n = p|n p) :

1. ω 0 (n) = 0 and ω 3 (n) = 0 or 1. 2. If ω 3 (n) = 1 then ℓ(n) + ℓ(n) ≡ 0 or 1 (mod 3). 3. If ω 3 (n) = 0 and ω 1 (n) + ℓ(n) -ℓ(n) is even then 2ℓ(n) -ℓ(n) ≡ 2 or 3 or 4 or 5 (mod 6). 4. If ω 3 (n) = 0 and ω 1 (n) + ℓ(n) -ℓ(n) is odd then 2ℓ(n) -ℓ(n) ≡ 0 or 4 (mod 6).
Remark 2. Point (b) of Theorem 0 can be improved in the following way : No element of A is a multiple of 31 2 . Indeed, from (1.10), we have for m odd, k ≥ 0 and τ ≥ 2,

31 τ mS A (31 τ m, k) = d | 31 τ m µ(d)σ A, 2 k 31 τ m d = d | 31m µ(d)σ A, 2 k 31 τ m d = d | m µ(d) σ A, 2 k 31 τ m d -σ A, 2 k 31 τ -1 m d .
Since 31 τ m d and 31 τ -1 m d are in the same orbit O(31) then (1.11) and (2.23) give σ(A,

2 k 31 τ m d ) ≡ σ(A, 2 k 31 τ -1 m d ) ≡ -5 (mod 2 k+1
), so that we get S A (31 τ m, k) ≡ 0 (mod 2 k+1 ). Hence, from (1.8), S A (31 τ m, k) = 0 and for all 0 ≤ h ≤ k and all τ ≥ 2, 2 h 31 τ m does not belong to A.

In view of stating Theorem 1 which will extend Theorem 0, we shall need some notation. The radical m of an odd integer m = 1, not divisible by 31 and free of prime factors belonging to P 0 will be written

m = p 1 . . . p ω 1 p ω 1 +1 . . . p ω 1 +ω 2 p ω 1 +ω 2 +1 . . . . . . p ω 1 +ω 2 +ω 3 +ω 4 +1 . . . p ω , (2.24) where ℓ(p i ) = j for ω 1 + ... + ω j-1 + 1 ≤ i ≤ ω 1 + ... + ω j , ω j = ω j (m) = ω j (m) and ω = ω(m) = ω(m) ≥ 1. We define the additive functions from Z \ 31Z into Z/12Z : α = α(m) = 2ω 5 -2ω 1 + ω 4 -ω 2 mod 12, (2.25)
a = a(m) = ω 5 -ω 1 + ω 2 -ω 4 mod 12.
(2.26)

Let (v i ) i∈Z be the periodic sequence of period 12 defined by

v i = 2 √ 3 cos(i π 6 ) if i is odd 2 cos(i π 6 ) if i is even. (2.27)
The values of (v i ) i∈Z are given by : i = 0 1 2 3 4 5 6 7 8 9 10 11

v i = 2 1 1 0 -1 -1 -2 -1 -1 0 1 1 Note that v i+6 = -v i , (2.28)
v i + v i+2 = v i+1 if i is odd 3v i+1 if i is even, (2.29) v 2i ≡ -2 i (mod 3) (2.30) and v i ≡ v i+3 ≡ v 2i (mod 2). (2.31)
From the U j 's (cf. (2.12) and (2.13)), we introduce the following 2-adic integers :

E i = 5 j=0 v i+2j U j , i ∈ Z, (2.32)
F i = 5 j=0 v i+4j U j , i ∈ Z, (2.33)
G = 5 j=0
(-1) j U j .

(2.34) From (2.28), we have

E i+6 = -E i , E i+12 = E i , F i+6 = -F i , F i+12 = F i . (2.35)
From (2.29), it follows that, if i is odd,

E i + E i+2 = E i+1 , F i + F i+2 = F i+1 , (2.36)
while, if i is even,

E i + E i+2 = 3E i+1 , F i + F i+2 = 3F i+1 , (2.37)
The values of these numbers are given in the following array :

Z Z mod 2 11 E 0 = 1 32 (11θ 5 -8θ 4 + 29θ 3 -124θ 2 + 500θ -256) E 1 = 1 16 (3θ 5 -2θ 4 + 9θ 3 -26θ 2 + 136θ -64) E 2 = 3E 1 -E 0 E 3 = 2E 1 -E 0 E 4 = 3E 1 -2E 0 E 5 = E 1 -E 0 F 0 = 1 32 (-3θ 5 -21θ 3 + 36θ 2 -36θ + 64) F 1 = 1 32 (-3θ 5 -4θ 4 -13θ 3 + 24θ 2 -28θ -64) F 2 = 3F 1 -F 0 F 3 = 2F 1 -F 0 F 4 = 3F 1 -2F 0 F 5 = F 1 -F 0 G = 1 4 (-θ 5 + θ 4 -θ 3 + 11θ 2 -34θ + 20) TABLE 1 Lemma 1. The polynomials (U j ) 0≤j≤5 (cf. (2.13)) form a basis of Q[θ]. The polynomials E 0 , E 1 , F 0 , F 1 , G, U 0 form an other basis of Q[θ].
For all i ′ s, E i and F i are linear combinations of respectively E 0 and E 1 and F 0 and F 1 .

Proof. With Maple, in the basis 1, θ, . . . , θ 5 , we compute determinant (U 0 , ..., U 5 ) = 1 1024 . From (2.32), (2.33) and (2.34), the determinant of (E 0 , E 1 , F 0 , F 1 , G, U 0 ) in the basis U 0 , U 1 , . . . , U 5 is equal to 12. The last point follows from (2.36) and (2.37).

We have Theorem 1. Let m = 1 be an odd integer not divisible by 31 with m of the form (2.24). Under the above notation and the convention

0 ω = 1 if ω = 0 0 if ω > 0, (2.38)
we have :

1) The 2-adic integer S(m) defined by (2.20) satisfies

mS(m) = 2 ω 3 -1 3 ⌈ ω 2 +ω 4 2 -1⌉ E α-2ℓ(m) + 0 ω 3 2 3 ⌈ ω 2 -1⌉ F a-4ℓ(m) + 0 ω 2 +ω 4 3 2 ω-1 (-1) ℓ(m) G. (2.39)
2) The 2-adic integer S(31m) satisfies

S(31m) = -31 -1 S(m), (2.40)
where 31 -1 is the inverse of 31 in Z 2 . In particular, for all k ∈ {0, 1, 2, 3, 4}, we have

2 k m ∈ A ⇐⇒ 31 • 2 k m ∈ A, since the inverse of 31 modulo 2 k+1 is -1 for k ≤ 4.
Proof of Theorem 1, 1). From (2.22), we have

mS(m) = d | m µ(d)U ℓ( m d ) = d | m µ(d)U ℓ(m)-ℓ(d) .
(2.41) Further, (2.41) becomes

mS(m) = 5 j=0 T (m, j)U ℓ(m)-j = 5 j=0 T (m, ℓ(m) -j)U j , (2.42)
with

T (m, j) = T (m, j) = d|m, ℓ(d)≡j ( mod 6)
µ(d).

(2.43) Therefore (2.39) will follow from (2.42) and from the following lemma :

Lemma 2. The integer T (m, j) defined in (2.43) with the convention (2.38) and the definitions (2.18) and (2.24)-(2.27), for m = 1, is equal to

T (m, j) = 2 ω 3 -1 3 ⌈ ω 2 +ω 4 2 -1⌉ v α-2j + 0 ω 3 2 3 ⌈ ω 2 -1⌉ v a-4j +0 ω 2 +ω 4 (-1) j 3 2 ω-1 .
(2.44)

Proof. Let us introduce the polynomial

f (X) = (1 -X) ω 1 (1 -X 2 ) ω 2 ...(1 -X 5 ) ω 5 = ν≥0 f ν X ν . (2.45)
If the five signs were plus instead of minus, f (X) would be the generating function of the partitions in at most ω 1 parts equal to 1, ..., at most ω 5 parts equal to 5. More generally, the polynomial

f (X) = ω i=1 (1 + a i X b i ) = ν≥0 f ν X ν
is the generating function of

f ν = ǫ 1 ,...,ǫω∈{0,1}, ω i=1 ǫ i b i =ν ω i=1 a ǫ i i .
To the vector ǫ = (ǫ 1 , ..., ǫ ω ) ∈ F ω 2 , we associate

d = ω i=1 p ǫ i i , µ(d) = ω i=1 (-1) ǫ i , L(d) = ω i=1 ǫ i ℓ(p i )
where L is the arithmetic function defined by (2.17) and we get

f ν = d|m, L(d)=ν µ(d), (2.46)
Consequently, by setting ξ = exp(

f ν = 1 6 5 i=0 ξ -ij f (ξ i ) = 1 6 5 i=1 ξ -ij f (ξ i ) = 1 6 5 i=1 ξ -ij (1 -ξ i ) ω 1 (1 -ξ 2i ) ω 2 (1 -ξ 3i ) ω 3 (1 -ξ 4i ) ω 4 (1 -ξ 5i ) ω 5 . (2.47)
By observing that

1-ξ = ξ 5 , 1-ξ 2 = ̺ = √ 3(cos π 6 -i sin π 6 ), 1-ξ 3 = 2, 1-ξ 4 = ̺, 1-ξ 6 = 0,
the sum of the terms in i = 1 and i = 5 in (2.47), which are conjugate, is equal to

2 6 R(ξ -j ξ 5ω 1 ̺ ω 2 2 ω 3 ̺ ω 4 ξ ω 5 ) = 2 ω 3 3 √ 3 ω 2 +ω 4 cos π 6 (2ω 5 -2ω 1 + ω 4 -ω 2 -2j).
(2.48) Now, the contribution of the terms in i = 2 and i = 4 is

2 6 R(ξ -2j ̺ ω 1 ̺ ω 2 0 ω 3 ̺ ω 4 ̺ ω 5 ) = 0 ω 3 √ 3 ω 1 +ω 2 +ω 4 +ω 5 3 cos π 6 (ω 2 + ω 5 -ω 1 -ω 4 -4j) = 0 ω 3 √ 3 ω 3 cos π 6 (ω 2 + ω 5 -ω 1 -ω 4 -4j) (2.49)
Finally, the term corresponding to i = 3 in (2.47) is equal to 1 6 (-1) j 2 ω 1 0 ω 2 2 ω 3 0 ω 4 2 ω 5 = 0 ω 2 +ω 4 (-1) j 6 2 ω 1 +ω 3 +ω 5 = 0 ω 2 +ω 4 (-1) j 6 2 ω . (2.50)

Consequently, by using our notation (2.24)-(2.26), (2.47) becomes

T (m, j) = 2 ω 3 3 √ 3 ω 2 +ω 4 cos π 6 (α -2j) + 0 ω 3 √ 3 ω 3 cos π 6 (a -4j) +0 ω 2 +ω 4 (-1) j 6 2 ω . (2.51)
Observing that α -2j has the same parity than ω 2 + ω 4 and similarly for a -4j and ω (when ω 0 = ω 3 = 0), via (2.27), we get (2.44).

Proof of Theorem 1, 2). For all k ≥ 0, from (1.10), we have

31mS A (31m, k) = d | 31m µ(d)σ(A, 31 • 2 k m d ) = d | 31m µ(d)σ(A, 31 • 2 k m d ) = d | m µ(d)σ(A, 31 • 2 k m d ) - d | m µ(d)σ(A, 2 k m d ) = d | m µ(d)σ(A, 31 • 2 k m d ) -mS A (m, k). (2.52) Since for all d dividing m, 31•2 k m d ∈ O(31) then, from (2.23), σ(A, 31•2 k m d ) ≡ σ(A, 31 • 2 k ) ≡ -5 (mod 2 k+1 ), so that (2.52) gives 31mS A (31m, k) + mS A (m, k) ≡ -5 d | m µ(d) (mod 2 k+1 ).
(2.53)

Since m = 1, 31mS A (31m, k) + mS A (m, k) ≡ 0 (mod 2 k+1
). Recalling that m is odd, by using (2.20), (2.21) and their similar for S(31m), we obtain the desired result.

Elements of the set

A = A(1 + z + z 3 + z 4 + z 5 ).
In this section, we will determine the elements of the set A of the form n = 2 k 31 τ m, where m = 1 satisfies (2.24) and τ ∈ {0, 1}, since from Remark 2, 2 k 31 τ m / ∈ A for all τ ≥ 2 . The elements of the set A(1 + z + z 3 + z 4 + z 5 ) of the form 31 τ 2 k , τ = 0 or 1, were shown in [START_REF] Baccar | Sets with even partition functions and 2-adic integers[END_REF] to be solutions of 2-adic equations. More precisely, the following was proved in that paper.

1) The elements of the set A(1 + z + z 3 + z 4 + z 5 ) of the form 2 k , k ≥ 0, are given by the 2-adic solution k≥0 χ(A, 2 k ) 2 k = S(1) = U 0 = 1 + 2 + 2 2 + 2 3 + 2 4 + 2 7 + 2 10 + 2 11 + ... of the equation y 6 -y 5 + 3y 4 -11y 3 + 44y 2 -36y + 32 = 0. Note that S(1) = U 0 follows from (2.22).

2) The elements of the set A(1 + z + z 3 + z 4 + z 5 ) of the form 31 • 2 k , k ≥ 0, are given by the solution Then, if 31 does not divide m, we have

γ(31m) = γ(m). (3.2)
Let us assume now that m is coprime with 31. We shall use the quantities ω i = ω i (m) defined by (2.18), ℓ(m), α = α(m), a = a(m) defined by (2.15), (2.25) and (2.26),

α ′ = α ′ (m) = α-2ℓ(m) mod 12 = 2ω 5 -2ω 1 +ω 4 -ω 2 -2ℓ(m) mod 12, (3.3) a ′ = a ′ (m) = a -4ℓ(m) mod 12 = ω 5 -ω 1 + ω 2 -ω 4 -4ℓ(m) mod 12, (3.4)
t = t(m) = ω 1 + ω 5 + ω 2 + ω 4 2 -1 - ω 2 + ω 4 2 -1 = ⌈ ω 1 +ω 5 2 ⌉ if ω 1 + ω 5 ≡ ω 2 + ω 4 ≡ 1 (mod 2) ⌈ ω 1 +ω 5 2 -1⌉ if not.
(3.5)

We have : (i) if ω 3 = 0 and ω 2 + ω 4 = 0, the value of γ = γ(m) is given by

γ =    ω 3 -1 if α ′ ≡ 0, 1, 3, 4 (mod 6) ω 3 if α ′ ≡ 2 (mod 6) ω 3 + 2 if α ′ ≡ 5 (mod 6). (ii) If ω 2 + ω 4 = 0 and ω 3 ≥ 1, we set α ′′ = α ′ + 6ℓ(m) mod 12 and δ(i) = v 2 (E i + 2 v 2 (E i ) G) and we have if ω 1 + ω 5 < v 2 (E α ′′ ), then γ = ω 3 -1 + ω 1 + ω 5 , if ω 1 + ω 5 = v 2 (E α ′′ ), then γ = ω 3 -1 + δ(α ′′ ), if ω 1 + ω 5 > v 2 (E α ′′ ), then γ = ω 3 -1 + v 2 (E α ′′ ).
(iii) If ω 3 = 0 and ω 2 + ω 4 = 0, we have

γ = -1 + v 2 (E α ′ + 3 t F a ′ ).
(iv) If ω 3 = ω 2 = ω 4 = 0 and ω 1 + ω 5 = 0, we have

γ = -1 + v 2 (E α ′ + 3 t F a ′ + 2 ω 1 +ω 5 (-1) ℓ(m) G).
Proof. We shall prove that S(m) = 0 in each of the four cases above. Assuming S(m) = 0, it follows from Theorem 1, 2) that S(31m) = 0 and that γ(31m) = γ(m), which sets (3.2).

Proof of Theorem 2 (i).

In this case, formula (2.39) reduces to

mS(m) = 2 ω 3 -1 3 ⌈ ω 2 +ω 4 2 -1⌉ E α ′ .
Since E α ′ = 0, S(m) does not vanish ; we have

γ = v 2 (S(m)) = ω 3 -1 + v 2 (E α ′ )
and the result follows from the values of E α ′ modulo 2 11 given in Table 1.

Proof of Theorem 2 (ii).

If ω 2 + ω 4 = 0 and ω 3 = 0, formula (2.39) becomes (since, cf. (2.35),

E i+6 = -E i holds) mS(m) = 2 ω 3 -1 3 E α ′ + 2 ω 1 +ω 5 (-1) ℓ(m) G = (-1) ℓ(m) 2 ω 3 -1 3 E α ′′ + 2 ω 1 +ω 5 G .
As displaid in Table 1, E i is a linear combination of E 0 and E 1 so that, from Lemma 1, S(m) does not vanish and γ = ω 3 -1+v 2 (E α ′′ + 2 ω 1 +ω 5 G), whence the result. The values of v 2 (E i ) and δ(i) calculated from Table 1 are given below. i 0 1 2 3 4 5 6 7 8 9 10 11 v 2 (E i ) 0 0 1 0 0 3 0 0 1 0 0 3

δ(i) 1 1 2 1 1 8 2 2 4 2 2 4
Proof of Theorem 2 (iii). If ω 3 = 0 and ω 2 + ω 4 = 0 it follows, from (2.39) and the definition of t above, that

mS(m) = 1 2 3 ⌈ ω 2 +ω 4 2 -1⌉ (E α ′ + 3 t F a ′ ).
But E i and F i are non-zero linear combinations of, respectively, E 0 and E 1 and F 0 and

F 1 ; by Lemma 1, E α ′ + 3 t F a ′ does not vanish and γ = -1 + v 2 (E α ′ + 3 t F a ′ ).
Proof of Theorem 2 (iv). If ω 3 = ω 2 = ω 4 = 0 and m = 1, formula (2.39) gives

mS(m) = 1 6 E α ′ + 3 t F a ′ + 2 ω 1 +ω 5 (-1) ℓ(m) G .
From Lemma 1, we obtain E α ′ + 3 t F a ′ + 2 ω 1 +ω 5 (-1) ℓ(m) G = 0, which implies S(m) = 0 and γ = -1

+ v 2 E α ′ + 3 t F a ′ + 2 ω 1 +ω 5 (-1) ℓ(m) G . Theorem 3.
Let m be an odd integer satisfying m = 1, (m, 31) = 1, and with m of the form (2.24). Let γ = γ(m) as defined in Theorem 2 and Z(m) be the odd part of the right hand-side of (2.39), so that mS(m) = 2 γ(m) Z(m).

(3.6)

(i) If k < γ, then 2 k m / ∈ A and 2 k 31m / ∈ A. (ii) If k = γ, then 2 k m ∈ A and 2 k 31m ∈ A. (iii) If k = γ + r, r ≥ 1, then we set S r = {2 r + 1, 2 r + 3, ..., 2 r+1 -1}
and we have

2 γ+r m ∈ A ⇐⇒ ∃ l ∈ S r , m ≡ l -1 Z(m) (mod 2 r+1 ), 2 γ+r 31m ∈ A ⇐⇒ ∃ l ∈ S r , m ≡ -(31l) -1 Z(m) (mod 2 r+1 ).
Proof of Theorem 3, (i). We remind that m is odd and (cf. 2.21) So that, by using Theorem 3, (i) and (1.8), we obtain

S(m) ≡ S A (m, k)(mod 2 k+1 ). It is obvious from (3.6) that if γ > k then S A (m, k) ≡ 0(mod 2 k+1 ). So that from (1.8), S A (m, k) = 0 and 2 h m ∈ A, for all h, 0 ≤ h ≤ k. To prove that 2 k 31m / ∈ A, it
2 k mχ(A, 2 k m) ≡ 2 k Z(m)(mod 2 k+1 ).
Since both m and Z(m) are odd, we get χ(A, 2 k m) ≡ 1( mod 2), which shows that 2 k m ∈ A. Once again, to prove that 2 k 31m ∈ A, it suffices to use this last result and (2.40) modulo 2 k+1 .

Proof of Theorem 3, (iii). Let us set

k = γ + r, r ≥ 1. (3.6) and (2.21) give mS A (m, k) ≡ 2 γ Z(m)(mod 2 γ+r+1 ). (3.7)
So that, by using Theorem 3, (i) and (ii), we get

m(2 γ + 2 γ+1 χ(A, 2 γ+1 m) + . . . + 2 γ+r χ(A, 2 γ+r m)) ≡ 2 γ Z(m)(mod 2 γ+r+1 ),
which reduces to

m(1 + 2χ(A, 2 γ+1 m) + . . . + 2 r χ(A, 2 γ+r m)) ≡ Z(m)(mod 2 r+1 ).
By observing that 2 γ+r m ∈ A if and only if l = 1 + 2χ(A, 2 γ+1 m) + . . . + 2 r χ(A, 2 γ+r m) is an odd integer in S r , we obtain

2 γ+r m ∈ A ⇐⇒ m ≡ l -1 Z(m) (mod 2 r+1 ), l ∈ S r .
To prove the similar result for 2 γ+r 31m, one uses the same method and (2.40) modulo 2 k+1 .

4 The counting function.

In Theorem 4 below, we will determine an asymptotic estimate to the counting function A(x) (cf. (1.2)) of the set A = A(1 + z + z 3 + z 4 + z 5 ). The following lemmas will be needed. Lemma 3. Let K be any positive integer and x ≥ 1 be any real number. We have

| {n ≤ x : gcd(n, K) = 1} |≤ 7 ϕ(K) K x,
where ϕ is the Euler function.

Proof. This is a classical result from sieve theory : see Theorems 3 -5 of [START_REF] Halberstam | Sieve methods[END_REF]. Lemma 4. (Mertens's formula) Let θ and η be two positive coprime integers. There exists an absolute constant C 1 such that, for all x > 1,

π(x; θ, η) = p≤x, p≡θ( mod η) (1 - 1 p ) ≤ C 1 (log x) 1 ϕ(η) .
Proof. For θ and η fixed, Mertens's formula follows from the Prime Number Theorem in arithmetic progressions. It is proved in [START_REF] Ben Saïd | On the counting function of the sets of parts such that the partition function takes even values for n large enough[END_REF] that the constant C 1 is absolute. Lemma 5. For i ∈ {2, 3, 4}, let

K i = K i (x) = p≤x, ℓ(p)∈{0,i} p = p≤x, p∈P 0 ∪Pi p,
where ℓ, P 0 and P i are defined by (2.15)-(2.16). Then for x large enough,

| {n : 1 ≤ n ≤ x, gcd(n, K i ) = 1} |= O x (log x) 1 3
.

Proof. By Lemma 3 and (2.16), we have

| {n : n ≤ x, gcd(n, K i ) = 1} |≤ 7x ϕ(K i ) K i = 7x 0≤j≤4, τ ∈{0,i} p≤x, p≡2 j 3 τ ( mod 31) (1 - 1 p ).
So that by Lemma 4, for all i ∈ {2, 3, 4} and x large enough,

| {n : n ≤ x, gcd(n, K i ) = 1} |≤ 7C 10 1 x (log x) 10 ϕ(31) = O x (log x) 1 3
. Lemma 6. Let r, u ∈ N 0 , ℓ and α ′ be the functions defined by (2.15) and (3.3), ω j be the additive function given by (2.18). We take ξ to be a Dirichlet character modulo 2 r+1 with ξ 0 as principal character and we let ̺ be the completely multiplicative function defined on primes p by

̺(p) = 0 if ℓ(p) = 0 or p = 31 1 otherwise. (4.1)
If y and z are respectively some 2 u -th and 12-th roots of unity in C, and if x is a real number > 1, we set

S y,z,ξ (x) = 2 ω 3 (n) n≤x ̺(n)ξ(n)y ω 2 (n)+ω 4 (n) z α ′ (n) . (4.2)
Then, when x tends to infinity, we have

• If ξ = ξ 0 , S y,z,ξ (x) = O x log log x (log x) 2 . (4.3)
• If ξ = ξ 0 , S y,z,ξ 0 (x) = x (log x) 1-fy,z(1) H y,z,ξ 0 (1)C y,z Γ (f y,z (1)) + O log log x log x , (4.4)
where Γ is the Euler gamma function,

f y,z (s) = 1 ϕ(31) 1≤j≤5 p, ℓ(p)=j g j,y,z (s), (4.5)
g 1,y,z (s) = z 8 , g 2,y,z (s) = yz 7 , g 3,y,z (s) = z 6 2 s , g 4,y,z (s) = yz 5 , g 5,y,z (s) = z 4 , (4.6)
H y,z,ξ (s) = 1≤j≤5 p, ℓ(p)=j 1 + g j,y,z (s)ξ(p) p s -z -2j ξ(p) 1 - ξ(p) p s g j,y,z (s)
,

C y,z = 1≤j≤5    p, ℓ(p)=j (1 - 1 p ) -g j,y,z (4.7)
p

(1 - 1 p ) g j,y,z (1) 30 
  . (4.8)
Proof. The evaluation of such sums is based, as we know, on the Selberg-Delange method. In [START_REF] Saïd | Sur une application de la formule de Selberg-Delange[END_REF], one finds an application towards direct results on such problems. In our case, to apply Theorem 1 of that paper, one should start with expanding, for complex number s with Rs > 1, the Dirichlet series

F y,z,ξ (s) = n≥1 ̺(n)ξ(n)y ω 2 (n)+ω 4 (n) z α ′ (n) (2 ω 3 (n) n) s
in an Euler product given by

F y,z,ξ (s) = 1≤j≤5 p, ℓ(p)=j 1 + ∞ m=1 ξ(p m )y ω 2 (p m )+ω 4 (p m ) z α ′ (p m ) (2 ω 3 (p m ) p m ) s = 1≤j≤5 p, ℓ(p)=j 1 + g j,y,z (s)ξ(p) p s -z -2j ξ(p) ,
which can be written

F y,z,ξ (s) = H y,z,ξ (s) 1≤j≤5 p, ℓ(p)=j 1 - ξ(p) p s -g j,y,z (s)
, where g j,y,z (s) and H y,z,ξ (s) are defined by (4.6) and (4.7). To complete the proof of Lemma 6, one has to show that H y,z,ξ (s) is holomorphic for Rs > 1 2 and, for y and z fixed, that H y,z,ξ (s) is bounded for Rs ≥ σ 0 > 1 2 , which can be done by adapting the method given in [START_REF] Saïd | Sur une application de la formule de Selberg-Delange[END_REF] (Preuve du Theorème 2, p. 235). Lemma 7. We keep the above notation and we let G be the set of integers of the form n = 2 ω 3 (m) m with the following conditions :

-m odd and gcd(m, 31) = 1, -m = m 1 m 2 m 3 m 4 m 5 , where all prime factors p of m i satisfy ℓ(p) = i.

If G(x) is the counting function of the set G then, when x tends to infinity,

G(x) = Cx (log x) 1/4 1 + O log log x log x , (4.9)
where Proof of Lemma 7. We apply Lemma 6 with y = z = 1, ξ = ξ 0 and remark that G(x) = S 1,1,ξ 0 (x). By observing that (1

C = H 1,1,ξ 0 (1)C 1,1 Γ (f 1,1 (1)
+ 1 p-1 )(1 -1 p ) = 1, we have H 1,1,ξ 0 (1) = p∈P 3 1 + 1 2(p -1) 1 - 1 p 1 2 = p∈P 3 1 - 1 2p 1 - 1 p -1 2 ≍ 1.000479390466, C 1,1 = lim x→∞ p∈P 1 ∪P 2 ∪P 4 ∪P 5 , p≤x 1 - 1 p -1 p∈P 3 , p≤x 1 - 1 p -1 2 p≤x 1 - 1 p 3 4
≍ 0.75410767606.

The numerical value of the above Eulerian products has been computed by the classical method already used and described in [START_REF] Saïd | Sur une application de la formule de Selberg-Delange[END_REF]. Since Γ (f 1,1 (1)) = Γ( 3 4 ) = 1.225416702465..., we get (4.10). Lemma 8. We keep the notation introduced in Lemmas 6 and 7. If (y, z) ∈ {(1, 1), (-1, -1)}, we have

S y,z,ξ 0 (x) = C x (log x) 1/4 1 + O log log x log x , (4.11)
while, if (y, z, ξ) / ∈ {(1, 1, ξ 0 ), (-1, -1, ξ 0 )}, we have

S y,z,ξ (x) = O r x (log x) 1/4+2 -2u-3 .
(4.12)

Proof. For y = z = 1, Formula (4.11) follows from Lemma 7. For y = z = -1 (which does not occur for u = 0), it follows from (4.4) and by observing that the values of g j,y,z (s), f y,z (s), H y,z,ξ (s), C y,z do not change when replacing y by -y and z by -z. Theorem 4. Let A = A(1 + z + z 3 + z 4 + z 5 ) be the set given by (1.3) and A(x) be its counting function. When x → ∞, we have

A(x) ∼ κ x (log x) 1 4
, where κ = 74 31 C = 1.469696766... and C is the constant of Lemma 7 defined by (4.10).

Proof of Theorem 4. Let us define the sets A 1 , A 2 , A 3 and A 4 containing the elements n = 2 k m (m odd) of A with the restrictions :

A 1 : ω 3 (m) = 0 and ω 2 (m) + ω 4 (m) = 0 A 2 : ω 3 (m) = 0 and ω 2 (m) = ω 4 (m) = 0 A 3 : ω 3 (m) = 0 and ω 2 (m) + ω 4 (m) = 0 A 4 : ω 2 (m) = ω 3 (m) = ω 4 (m) = 0.
We have

A(x) = A 1 (x) + A 2 (x) + A 3 (x) + A 4 (x). (4.14)
Further, for i = 2, 3, 4, it follows from Lemma 5 that

A i (x) = O x (log x) 1 3
and therefore

A(x) = A 1 (x) + O x (log x) 1 3
.

(4.15)

Now, we split A 1 in two parts B and B by putting in B the elements n ∈ A 1 which are coprime with 31 and in B the elements n ∈ A 1 which are multiples of 31. Let us recall that, from Remark 2, no element of A is a multiple of 31 2 . Therefore,

A 1 (x) = B(x) + B(x) (4.16) with B(x) = n=2 k m∈A 1 , n≤x ρ(m), B(x) = n=2 k 31m∈A 1 , n≤x ρ(m). ( 4

.17)

Let us consider B(x) ; the case of B will be similar. We define

ν i = v 2 (E i ) -1 =    -1 if i ≡ 0, 1,
E i = 2 -1-ν i E i . (4.19) In view of Theorem 2 (i), if i = α ′ (m) mod 12 then γ(m) -ω 3 (m) = ν i . (4.20)
Further, an element n = 2 k m (m odd) belonging to A 1 is said of index r ≥ 0 if k = γ(m) + r. For r ≥ 0 and 0 ≤ i ≤ 11,

T (i) r (x) = n=2 γ(m)+r m∈A 1 , n≤x α ′ (m)≡i (mod 12) ρ(m) = n=2 γ(m)+r m∈A 1 , 2 ω 3 (m) m≤2 -r-ν i x α ′ (m)≡i (mod 12)
ρ(m) (4.21) will count the number of elements of A 1 up to x of index r and satisfying α ′ (m) ≡ i (mod 12), so that

B(x) = r≥0 11 i=0 T (i) r (x). ( 4
T (i) r (x) ≤ G(2 1-r x) (4.24)
with G defined in Lemma 7. Moreover, from Lemma 7, there exists an absolute constant K such that, for x ≥ 3,

G(x) ≤ K x (log x) 1 4 • (4.25) Now, let R be a large but fixed integer ; R ′ is defined in terms of x by 2 R ′ -1 ≤ √ x < 2 R ′ and R" = log x log 2 . Since T (i) r (x) is a non-negative integer, (4.23) implies that T (i) r (x) = 0 for r > R". If x is large enough, R < R ′ < R ′′ holds. Setting B R (x) = R r=0 11 i=0 T (i) r (x), ( 4
T (i) r (x), S" = R" r=R ′ +1 11 i=0 T (i) r (x).
The definition of R ′ and (4.23) yield

S" ≤ R" r=R ′ +1 x 2 r ≤ ∞ r=R ′ +1 x 2 r = x 2 R ′ ≤ √ x,
while (4.24), (4.25) and the definition of R ′ give

S ′ ≤ R ′ r=R+1 G x 2 r-1 ≤ R ′ r=R+1 2Kx 2 r log x 2 R ′ -1 1 4 ≤ 2 5 4 Kx (log x) 1 4 R ′ r=R+1 1 2 r ≤ 3Kx 2 R (log x) 1 4
, so that, for x large enough, we have

0 ≤ B(x) -B R (x) ≤ √ x + 3Kx 2 R (log x) 1 4
• (4.27)

We now have to evaluate T (i) r (x) ; we shall distinguish two cases, r = 0 and r ≥ 1.

Calculation of T

(i) 0 (x). From (4.21), we have

T (i) 0 (x) = n=2 γ(m) m∈A 1 , n≤x α ′ (m)≡i (mod 12) ρ(m) = n=2 γ(m) m∈A, n≤x, ω 3 =0, ω 2 +ω 4 =0 α ′ (m)≡i (mod 12)
ρ(m).

From Theorem 3, we know that 2 γ(m) m ∈ A. Hence,

T (i) 0 (x) = 2 γ(m) m≤x, ω 3 =0, ω 2 +ω 4 =0 α ′ (m)≡i (mod 12) ρ(m),
which, by use of (4.20), gives

T (i) 0 (x) = 2 ω 3 (m) m≤2 -ν i x, ω 3 =0, ω 2 +ω 4 =0 α ′ (m)≡i (mod 12)
ρ(m).

But, at the cost of an error term O

x

(log x) 1 3
, Lemma 5 allows us to remove the conditions ω 3 = 0, ω 2 +ω 4 = 0, and to get from the second part of Lemma 9, From (4.21), it follows that ρ(m). Now, by Theorem 3, we know that 2 γ(m)+r m belongs to A if there is some l ∈ S r = {2 r + 1, ..., 2 r+1 -1} such that m ≡ l -1 Z(m) mod 2 r+1 . Note that the order of 3 modulo 2 r+1 is 2 r-1 if r ≥ 2 and 2 r if r = 1. We choose u = r + 1 so that ω 2 + ω 4 ≡ λ (mod 2 r+1 ) implies 3 ⌈ λ 2 -1⌉ ≡ 3 ⌈ ω 2 +ω 4 2 -1⌉ (mod 2 r+1 ). Therefore, we have

T (i) 0 (x) = G i,0,0,0,1 x 2 ν i + O x (log x) 1 3 = C 12 x 2 ν i (log x) 1 4 1 + O 1 (log x) 1/12 . ( 4
T (i) r (x) =
T (i) r (x) = l∈Sr 2 r+1 -1 λ=0
2 ω 3 (m) m≤2 -ν i -r x, ω 3 =0, ω 2 +ω 4 =0 α ′ (m)≡i (mod 12), ω 2 +ω 4 ≡λ (mod 2 r+1 )

m≡l -1 3 ⌈ λ 2 -1⌉ E i (mod 2 r+1 ) ρ(m).
As in the case r = 0, we can remove the conditions ω 3 = 0 and ω 2 + ω 4 = 0 in the last sum by adding a O

x (log x) Cx (log x)

1 4 3 2 - 1 2 R 1 + O 1 (log x) 2 -2R-5
.

By making R going to infinity, the above equality together with (4.27) show that B(x) ∼ 37 16

Cx (log x)

1 4
, x → ∞. (4.30)

In a similar way, we can show that B(x) defined in (4.17

Numerical computation of A(x).

There are three ways to compute A(x). The first one uses the definition of A and simultaneously calculates the number of partitions p(A, n) for n ≤ x ; it is rather slow. The second one is based on the relation (1.10) and the congruences (2.19) and (2.23) satisfied by σ(A, n). The third one calculates ω j (n), 0 ≤ j ≤ 5, in view of applying Theorem 1. The two last methods can be encoded in a sieving process

The following table displays the values of A(x), A 1 (x), ..., A 4 (x) as defined in (4.14) and also c(x) = A(x)(log x)

1 4
x , c 1 (x) = A 1 (x)(log x) 1 4

x .

It seems that c(x) and c 1 (x) converge very slowly to κ = 1.469696766 . . ., which is impossible to guess from the table.

x A(x) c(x) A 1 (x) c 1 (x) A 2 (x) A 3 (x) A 4 (x)

Thanks

We are pleased to thank A. Sárközy who first considered the sets A's such that the number of partitions p(A, n) is even for n large enough for his interest in our work and X. Roblot for valuable discussions about 2-adic numbers.

p

  ) which, by Möbius inversion formula, gives mS A (m, k) = denotes the radical of m with 1 = 1.

  and not a multiple of 31, then m ∈ A if and only if 31m ∈ A.

Theorem 2 .

 2 k≥0 χ(A, 31 • 2 k ) 2 k = S(31) = y = 2 2 + 2 5 + 2 11 + ... of the equation 31 5 y 6 + 31 5 y 5 + 13 • 31 4 y 4 + 91 • 31 3 y 3 + 364 • 31 2 y 2 + 796 • 31y + 752 = 0, since, from (2.53) with m = 1, we have 31S(31) = -5 -U 0 , so that S(31) = 5 + U 0 1 -32 = (1 + 4 + U 0 )(1 + 2 5 + 2 10 + ...) = 2 2 + 2 5 + 2 11 + ... Let m = 1 be an odd integer not divisible by any prime p ∈ P 0 (cf. (2.16)) neither by 31 2 . Then the sum S(m) defined by (2.20) does not vanish. So we may introduce the 2-adic valuation of S(m) : γ = γ(m) = v 2 (S(m)).(3.1)

  suffices to use this last result and (2.40) modulo 2 k+1 . Proof of Theorem 3, (ii). If γ = k then the same arguments as above show that mS A (m, k) ≡ 2 k Z(m)(mod 2 k+1 ).

H 1 , 1 ,

 11 ξ 0 (1), C 1,1 and f 1,1 (1) are defined by (4.7),(4.8) and (4.5).

Since ν i ≥ - 1 ,

 1 .22) Since γ(m) ≥ 0, from the first equality in (4.21), each n counted in T (i) r (x) is a multiple of 2 r , hence the trivial upper bound the second equality in (4.21) implies 11 i=0

  .26) from (4.22), we have B(x) -B R (x) = S ′ + S",

r 2 -

 2 (x) for r ≥ 1. Under the conditions ω 3 = 0 and ω 2 + ω 4 = 0, from (3.6), (2.39), (3.3), (4.19) and (4.20), we get Z(m) = 3 ⌈ ω 2 +ω 4 1⌉ E α ′ (m) .

  n=2 γ(m)+r m∈A, n≤x, ω 3 =0, ω 2 +ω 4 =0 α ′ (m)≡i (mod 12)

  (4.16) and (4.15), completes the proof of Theorem 4 with

  4 -11y 3 + 44y 2 -36y + 32.Note that R(y)5 is the resultant in z of φ 31 (z) = 1 + z + ... + z 30 and y + z + z 2 + z 4 + z 8 + z 16 .

	Let us set
				θ = U 0 = 1 + 2 + 2 2 + 2 3 + 2 4 + 2 7 + 2 10 + ...
	It turns out that the Galois group of R(y) is cyclic of order 6 and therefore the
	other roots U 1 , ..., U 5 of R(y) are polynomials in θ. With Maple, by factorizing
	R(y) on Q[θ] and using the values of u 11,j , we get
					U 0 = θ ≡ 1183 (mod 2 11 )
		U 1 =	1 32	(3θ 5 + 5θ 3 -36θ 2 + 84θ) ≡ 1598 (mod 2 11 )
		U 2 =	1 32	(-3θ 5 -5θ 3 + 20θ 2 -100θ) ≡ 1554 (mod 2 11 )
		U 3 =	1 32	(-θ 5 -7θ 3 + 12θ 2 -44θ + 32) ≡ 845 (mod 2 11 )
	U 4 =	1 32	(-θ 5 + 4θ 4 + θ 3 + 24θ 2 -68θ + 96) ≡ 264 (mod 2 11 )
	U 5 =	1 16	(θ 5 -2θ 4 + 3θ 3 -10θ 2 + 48θ -48) ≡ 701 (mod 2 11 ).	(2.13)

Let us define M y,z = ℜ(f y,z (1)) = 1 6 ℜ(z 6 (z 2 + z -2 + 1 2 + y(z + z -1 ))).

When ξ = ξ 0 , (4.3) implies (4.12) while, if ξ = ξ 0 , it follows from (4.4) and from the inequality to be proved

To show (4.13), let us first recall that z is a twelfth root of unity.

for all u ≥ 0, which proves (4.13).

If z = 1 and y = 1 (which implies u ≥ 1), we have

and

• If z = -1 and y = -1, (4.13) follows from the preceding case by observing that f y,z (1) = f -y,-z (1), which completes the proof of (4.13).

Lemma 9. Let G be the set defined in Lemma 7, ω j and α ′ be the functions given by (2.18) and (3.3). For 0 ≤ j ≤ 11, r, u, λ, t ∈ N 0 such that t is odd, we let G j,r,u,λ,t be the set of integers n = 2 ω 3 (m) m in G with the following conditions :

-

). If ρ is the function given by (4.1), the counting function G j,r,u,λ,t (x) of the set G j,r,u,λ,t is equal to

If u ≥ 1 and λ ≡ j (mod 2), G j,r,u,λ,t is empty while, if λ ≡ j (mod 2), when x tends to infinity, we have

where C is the constant given by (4.10).

If u = 0, then

By using the relations of orthogonality :

In the above triple sums, the main contribution comes from S 1,1,ξ 0 (x) and S -1,-1,ξ 0 (x), and the result follows from (4.11) and (4.12).

If u = 0, we have G j,r,0,0,t (x) = 1 12 • 2 r ξ mod 2 r+1 11 j 2 =0 ξ(t)µ -jj 2 S 1,µ j 2 ,ξ (x) and, again, the result follows from Lemma 8.