Discrete uniform limit law for additive functions on shifted primes

. The sufﬁcient and necessary conditions for a weak convergence of distributions of a set of strongly additive functions f x , x (cid:62) 2 , the arguments of which run through shifted primes, to the discrete uniform law are obtained. The case when f x ( p ) ∈ { 0 , 1 } for every prime p is considered.


Introduction
Let f x , x 2, be a set of strongly additive functions such that f x (p) ∈ {0, 1} for all primes p and all x 2. It follows from the strong additivity that for every positive integer n The problem of a weak convergence of distributions is of key importance in probabilistic number theory.There are interesting general conditions of convergence, classes of possible limit distributions, conditions of convergence to particular distributions.A detailed account of particular and general results you can find in the monographs [1,2,5].
In the articles [7,8,9], the case of the Poisson distribution as a limit law was considered.It was shown there that the Poisson law can occur as a limit one for the distributions: ν x p x, f x (p + 1) < u , ν x n x, f x (n) + g x (n + 1) < u , ν x p x, f x (p + 1) + g x (p + 2) < u .
In this work, we consider the weak convergence of distribution functions to the discrete uniform law where the parameter L ∈ N, L 2. Similarly as in [9], we use in the proofs the method of factorial moments and we have to restrict the behaviour of additive functions on large primes (see condition (H)).But the authors think that this condition is not necessary for the weak convergence to the discrete uniform distribution.Maybe, the problem could be solved applying the Kubilius model of probability spaces [1,2,5].But the large primes have to be overcome there as well.
Throughout the paper, we keep the following notation.The values of p, p 1 , p 2 , . . .mean prime numbers; c is an absolute positive constant not always the same.By the symbol ε(x) we denote values vanishing as x → ∞.The notation a b is equivalent to the inequality |a| cb.If the constant c, the constant included in , or the vanishing function ε(x) depend on a parameter a, we write c a , a , ε a (x).The notation F x (u) ⇒ F (u) means that the distribution functions F x (u) converge weakly to the distribution function F (u) as x → ∞.The superscript * at the signs of sum or maximum, * , max * , means that the summation or maximum is expanded over the primes p for which f x (p) = 1.The other notation is generally accepted or is later discussed in the text.for all γ ∈ (0, 1).The distributions ν x (f x (p + 1) < u) converge weakly to the limit discrete uniform law U(u, L) as x → ∞ if and only if L = 2 and

Main result and auxiliary lemmas
The proof of this main theorem is based on the following three lemmas on the limit behaviour of factorial moments of the distribution ν x (f x (p + 1) < u).
Lemma 1.Let f x , x 2, be a set of strongly additive functions such that f x (p) ∈ {0, 1} for all primes p.If distributions (1) converge weakly to some distribution function F (u) with a jump at the point u = 0 as x → ∞, then the quantities where g l is the lth factorial moment of the limit law. then According to this statement and equality (4) in the case of convergence of ν x (f x (p + 1) < u), we have that for each l ∈ {1, 2, . . .}.
, be a set of strongly additive functions such that f x (p) ∈ {0, 1} for all primes p and condition (H) hold.If distributions (1) converge weakly to the distribution F ξ of the random variable ξ with a finite support {0, 1, . . ., L − 1}, L 2, then there exists some constant D 2 such that Moreover, the characteristic function of the limit distribution F ξ is equal to From Theorem 1 we get the following example.

Proofs of lemmas
Proof of Lemma 1. Suppose that distribution functions (1) converge weakly to the limit distribution F (u) with the jump at the point u = 0. From the weak convergence we have that lim Using this estimate, it is proved in [9] (see inequality (10)) that According to this, http://www.mii.lt/NA for every k l + 2. Therefore, for every k l + 2.
Let us fix l ∈ N and choose K > l + 2. Using estimate (10), analogously as in [9], we get Taking the limit in the last equality as x tends to infinity and then as K tends to infinity, we obtain relation (4).Lemma 1 is proved.
Proof of Lemma 3. The proof of the lemma almost coincides with the proof of the necessity part of Corollary 4 from [6].From the conditions of the lemma we have that there is some k ∈ {0, 1, . . ., L − 1} such that Thus, from the inequality (see [3]) .
Assuming that lim sup Since for every pair i, j, Thus, the condition g L = 0 implies (8).The last condition (9) of the lemma now follows from (6), equality (8), and condition (H).Lemma 3 is proved.http://www.mii.lt/NA 4 Proof of Theorem 1 Necessity.Suppose that with parameter L 2. From Lemma 1 we obtain that for l = 1, 2, . . ., L−1.The values of g l are the factorial moments of the limit distribution.In the case of the uniform distribution, we get that From (12) we have that Thus, inequality (5) follows from inequality (11) with k = 0. We apply now Lemma 2. The values of g l are the factorial moments of the limit distribution.It is clear that for all l = 2, 3, . . .and all k = 1, 2, . . ., l − 1.In the particular case, Therefore, which implies L 5. Further, we examine separately the cases L = 2, 3, 4, 5.
Let L = 2.In this case, Using Lemma 3, we have that, for some D 2, lim sup Nonlinear Anal.Model.Control, 21(4):437-447 If κ = 1, then we obtain from relations above and condition (H) that there is only one case for large x and If κ = 0, then f x (p) = 0 for every fixed p and sufficiently large x.In this case, equality (13) cannot be satisfied.It follows that the case κ = 0 cannot occur.
Let L = 3.Then According to Lemma 3, we have that, for some fixed D 2, the following conditions hold: First, we suppose that κ = 2. From ( 14) and condition (H) we have that for large x and for some fixed primes p 1 < p 2 .Since the last equality is impossible for any pair of different primes p 1 , p 2 , then the case κ = 2 cannot occur.
From equality (15) it follows that the cases κ = 1, κ = 0 cannot occur as well.
Let L = 5.Then According to Lemma 3, there exists some D 2 for which It follows from equality (21) that κ cannot be equal to 0, 1, 2, and 3. Suppose κ = 4. Then from (20) we deduce that there exist primes p 1 < p 2 < p 3 < p 4 such that f x (p 1 ) = f x (p 2 ) = f x (p 3 ) = f x (p 4 ) = 1 for large x and So, the case κ = 4 is impossible as well.
Sufficiency.Assume that conditions (3) of the theorem together with additional condition (H) are satisfied.Estimate (5) follows from condition (3).Now using Lemma 2, we obtain that Taking the limit in the last equality, we conclude that But this is the characteristic function of the uniform distribution U(u, 2).So, the sufficiency of the theorem follows.Theorem 1 is now proved. http://www.mii.lt/NA

Theorem 1 .
Let f x , x 2, be a set of strongly additive functions.Assume that f x (p) ∈ {0, 1} for all prime numbers p and lim x→∞ http://www.mii.lt/NA

Lemma 2 .
(See [9, Lemma 2].)If a set of strongly additive functions f x satisfies the conditions of Theorem 1 and *

x→∞#
p d: f x (p) = 0 L, we get a contradiction to the condition g L = 0. Thus, lim sup x→∞ # p d: f x (p) = 0 L − 1 for each fixed positive integer d.Put a d = lim sup x→∞ # p d: f x (p) = 0 .The sequence a d (d 2) is integer-valued, non-decreasing, and bounded.Thus, there exists a positive integer D 2 such that lim sup x→∞ # p D: f x (p) = 0 = a d for d D. Since a d L − 1 for all positive integers d, we obtain condition (7).On the other hand, from the reasoning above it follows that lim x→∞ f x (p) = 0 for each fixed prime p > D. Consequently, we obtain that lim