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n 2. The Euler phi function : ϕ(n)= 1. k=1 gcd(Pk,n)=1

3. The number of distinct prime divisors of n : ω(n)= 1 . pP|n 1 if n =1 4. The Mobiuse function : µ(n)=  0 if p2|n for some prime p  (−1)ω(n) otherwise  5. number of positive divisors of n defined by : τ(n)= 1 . dP|n 6. sum of divisors function of n defined by : σ(n)= d . dP|n

Now ,if f,g : N −→ C are two arithmetic functions from the positive integers to the complex numbers, the Dirichlet convolution f ∗ g is a new arithmetic function defined by: n (f ∗ g)(n)= f(d)g( )= f(a)g(b) (3) d Xd|n abX=n where the sum extends over all positive divisors d of n , or equivalently over all distinct pairs (a,b) of positive integers whose product is n. In particular, we have (f ∗ g)(1) = f(1)g(1) ,(f ∗ g)(p)= f(1)g(p)+ f(p)g(1) for any prime p and for any power prime pm we have : m (f ∗ g)(pm)= f(pj)g(pm−j ) (4) Xj=0

This product occurs naturally in the study of Dirichlet series such as the Riemann zeta function. It describes the multiplication of two Dirichlet series in terms of their coefficients:

f ∗ g (n) f(n) g(n) = (5)  ns   ns  ns  nX≥1 nX≥1 nX≥1 with Riemann zeta function or is defined by : 1 ζ(s)= ns nX≥1

These functions are widely studied in the literature (see, e.g., [5, 6, 7]).

Now before to past to main result we need this propriety , if f and g are multiplicative function , then f ∗ g is multiplicative.

2 Main results

In this section we give the new result of Dirichlet product between derivative arithmetic and an arithmetic function multiplicative f , and we will give the relation between τ and the derivative arithmetic . Theorem 4. Given a multiplicative function f, and lets n and m two positive integers such that gcd(n,m)=1 , Then we have : (f ∗ δ)(nm)= Id ∗ f (n). f ∗ δ (m)+ Id ∗ f (m). f ∗ δ (n) (6)     2 Dirichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT

Proof. Lets n and m two positive integers such that gcd(n,m)=1, and let f an arithmetic function multiplicative , then we have : nm nm n m (f ∗ δ)(nm)= f δ(d)= f( )δ(d1d2)= f( )f( ) d1δ(d2)+ d2δ(d1) d d1d2 d1 d2   dX|nm  dX1|n dX1|n d2|m d2|m n m m n = d1f( )f( )δ(d2)+ d2f( )f( )δ(d1)  d1 d2 d2 d1  dX1|n d2|m n m m n = d1f( ) f( )δ(d2) + d2f( ) f( )δ(d1)  d1  d2   d2  d1  dX1|n dX2|m dX2|m dX1|n = Id ∗ f (n). f ∗ δ (m)+ Id ∗ f (m). f ∗ δ (n)    

s αi Lemma 5. For any n , if n = i=1 pi is the prime factorization of n, then : Q s f ∗ δ (pαi ) f ∗ δ (n)= Id ∗ f (n) i (7) Id ∗ f (pαi )   Xi=1 i 

α1 αs Proof. Let n a positive integer such that n = p1 ...ps and let f an arithmetic function , Then :

α1 αs f ∗ δ (n)= f ∗ δ (p1 ...ps )   α2 αs α1 α1 α2 αs = Id ∗ f (p2 ...ps ). f ∗ δ (p1 )+ Id ∗ f (p1 ). f ∗ δ (p2 ...ps )  f ∗ δ (pα1 )    = Id ∗ f (n). 1 + Id ∗ f (pα1 ). Id ∗ f (pα3 ...pαs ). f ∗ δ (pα2 )+ Id ∗ f (pα1 ) 1  3 s 2  1     + Id ∗ f (pα2 ). f ∗ δ (pα3 ...pαs ) 2 3 s    f ∗ δ (pα1 ) f ∗ δ (pα2 ) = Id ∗ f (n) 1 + Id ∗ f (n) 2 + Id ∗ f (pα1 ) Id ∗ f (pα2 )  1  2 α1  α2 α3 αs  + Id ∗ f (p1 ) Id ∗ f (p2 ) f ∗ δ (p3 ...ps ) .    . f ∗ δ (pα1 ) f ∗ δ (pα2 ) f ∗ δ (pαs ) = Id ∗ f (n) 1 + Id ∗ f (n) 2 + ... + Id ∗ f (n) s Id ∗ f (pα1 ) Id ∗ f (pα2 ) Id ∗ f (pαs )  1  2  s f ∗ δ (pα1 ) f ∗δ (pαs)  = Id ∗ f (n) 1 + ... + s  Id ∗ f (pα1 ) Id ∗ f (pαs )  1 s s f ∗δ (pαi )  = Id ∗ f (n) i Id ∗ f (pαi )  Xi=1 i 

α s f δ p i s αi ∗ ( i ) an other prof by induction on s that if n = i pi then (f ∗ δ)(n)= Id ∗ f (n) α . =1 Id∗f (p i ) Q  iP=1 i 

N s αi Proof. Consider n ∈ and express n = i=1 pi where all pi are distinct . α α 1 f∗δ (p i ) f∗δ (p 1 ) Q i α1 1 α1 where s =1 , it is clear that (f ∗ δ)(n)= Id ∗ f (n) α = Id ∗ f (p ) α = f ∗ δ (p1 ). Id∗f (p i ) Id∗f (p 1 )  iP=1 i  1    3 Dirichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT

s αi Assume that n = i=1 pi , then we have :

αs+1 Q αs+1 αs+1 id ∗ δ (n.ps+1 )= Id ∗ f (ps+1 ). f ∗ δ (n)+ Id ∗ f (n). f ∗ δ (ps+1 )    s f ∗δ (pαi )  f ∗ δ (pαs+1 ) = Id ∗ f (pαs+1 ). Id ∗ f (n) i + Id ∗ f (pαs+1 ). Id ∗ f (n) s+1 s+1 Id ∗ f (pαi ) s+1 Id ∗ f (pαs+1 )   Xi=1 i   s+1 s  αs+1  f ∗ δ (pαi ) f ∗ δ (p ) = Id ∗ f (n.pαs+1 ) i + Id ∗ f (n.pαs+1 ) s+1 s+1 Id ∗ f (pαi ) s+1 Id ∗ f (pαs+1 )  Xi=1 i  s+1 s  αs+1  f ∗ δ (pαi ) f ∗ δ (p ) = Id ∗ f (n.pαs+1 ) i + s+1 s+1  Id ∗ f (pαi ) Id ∗ f (pαs+1 )  Xi=1 i s+1   s+1 f ∗ δ (pαi ) = Id ∗ f (n.pαs+1 ) i s+1 Id ∗ f (pαi )  Xi=1 i 

Proposition 6. Let f a function arithmetic multiplicative , and δ the derivative arithmetic , then we have : 1 Id ∗ δ (n)= τ(n)δ(n) (8) 2  n Proof. Since (Id ∗ Id)(n)= d d = n 1= nτ(n). d|n d|n α P P α α j α−j j−1 α−j 1 α−1 and : (Id ∗ δ)(p )= δ(p )Id(p )= jp p = 2 α(α + 1)p . j=1 j=1 P P α1 αs Then for every a positive integer n such that n = p1 ...ps , we have : s Id ∗ δ (pαi ) Id ∗ δ (n)= Id ∗ Id (n) i Id ∗ Id (pαi )   Xi=1 i s 1 αi−1 αi(αi + 1)p = nτ(n) 2 i pαi τ(pαi ) Xi=1 i i s 1 αi−1 2 αi(αi + 1)pi = nτ(n) αi p (αi + 1) Xi=1 i s 1 α 1 = nτ(n) i = τ(n)δ(n) 2 pi 2 Xi=1

So by the proposition 6 , and the equality 5 we have this relation between arithmetic derivative and the function τ : δ(n) δ(n)τ(n) 2ζ(s − 1) = (9) ns ns nX≥1 nX≥1

Let defined the new function arithmetic called En-naoui function , by : 1 Φϕ(n)= n 1 − (10)  p Xp|n Then we have this equality related between 8 arithmetic function : Φ (n) B ∗ Id (n) µ ∗ δ (n)= ϕ(n) δ(n) − 2ω(n)+ B(n)+ ϕ + . (11)  n σ(n)   with B is the arithmetic function defined by : B(n)= αp. In next article i will prove this equality, just I need to α pP||n submit an article about this new function .

4 Dirichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT

3 Acknowledgements

I would like to thank S.Allo for the opportunity to collaborate on my research and for his continued mentorship, and friendship, and I’m indebted to the Professor Zouhaïr Mouayn to endorse me to submit my article here.I also extend my deepest appreciation to my family for opening their minds, hearts, and homes to me and my work.

References

[1] E. J. Barbeau. Remark on an arithmetic derivative, Canad. Math. Bull. 4 (1961), 117–122. [2] A. Buium. Arithmetic analogues of derivations, J. Algebra, 198 (1997), 290–299. [3] V. Ufnarovski and B. Åhlander. How to differentiate a number, J . Integer Seq., 6 , (2003). [4] M. Stay. Generalized number , J . Integer Seq., 8 , (2005). [5] T. M. Apostol. Introduction to Analytic , Springer-Verlag, New York , (1976). [6] M.Lewinter and J.Meyer. Elementary number theory with programming , Wiley , (2016). [7] Heng Huat Chan, Analytic number theory for undergraduates, World Scientific Publishing Co. Pte. Ltd. , (2009).

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