INVERSIVE CONGRUENTIAL GENERATOR WITH A VARIABLE SHIFT OF PSEUDORANDOM POINTS OVER THE COMPLEX PLANE Tran

Consider the generator of pseudorandom points on unit square produced by the inversive congruential recursion over the ring of Gaussian integers. Study the exponential sums on sequences of these points. AMS Subject Classification: 11K45, 11T71, 94A60, 11L07, 11T23


Introduction
Inversive congruential generator of pseudorandom numbers (PRN's) on the unit segment [0, 1) of real line arose as an alternative to D. Lemer's linear generator that didn't guarantee the "unpredictability" of elements of the generated sequence.It turned out that the non-linear congruential generators of the sequence of PRN's only may provide the unpredictability of the sequence of PRN's.
We say, that the sequence of real numbers {x n }, x n ∈ [0, 1), n = 0, 1, 2, . . .be the sequence of PRN's if it is generated by the determine algorithm and behaves like the sequence of implementations of random variables ξ 0 , ξ 1 , ξ 2 . . .that are uniformly distributed and statistically independent.Such definition of the sequence of PRN's is quiet reasonable to be apply in the modelling of stochastic processes and cryptography(for example, in forming the random key).Since every congruent sequence has a period that is not great than module of congruence, for application it is necessary to guarantee the pretty large period length.The one needs also an efficient software and hardware implementation for respective recursion.
In 1986 Eichenauer and Lehn [3] and then the Niederreiter [12][13] proposed the inversive congruential generator defined by recursion where y 0 ∈ Z * p , a, b ∈ Z, y −1 n is a solution of congruence y 0 x ≡ 1 (mod p) if (y n , p) = 1, or y −1 n = 0 if y n = 0.Under certain conditions, the recursion (1) generates the sequence {y n } (and therefore the sequence x n = yn p ∈ [0, 1)).It is clear that the sequence {y n } has a period τ ≤ p, and so for the applications it is necessary to choose a big prime number p.At the present time there are described the conditions when the period of its sequence is near to p (see, for example, Chou [1]).A big period of sequence of PRN's may be provide by recursion where p ≥ 2 is prime, m is natural, a, b, y 0 ∈ Z p m , (y 0 , p) = 1, y −1 n is a multiplicative inverse to y n (mod p m ).
Not less important for applications is to be able to build the sequences of pseudorandom points The present paper is concerned with studying the distribution of points of the unit square in C that are generated by recursion over the ring of Gaussian integers: where α, β, γ(n), z 0 ∈ Z[i], p is prime number, p ≡ 3 (mod 4).If γ(n) = Const, the generator (3) is essentially call the inversive generator with a variable shift.The selection of prime number p ≡ 3 (mod 4) comes from the fact that such prime rational numbers (and only one) be primes in the ring of Gaussian integers.
We construct two representations of z n : • in the form of polynomials in z 0 and z −1 0 modp m (with coefficients depending on n) and • in the form of polynomials in n (with coefficients depending on z 0 and z −1 0 modp m ).
These representations allow to obtain the non-trivial estimates of exponential sums on the elements of sequence {z n }.And, by virtue of Turan-Erdös-Koksma inequality [15], the non-trivial estimates of exponential sums on elements of sequence {z n } allow to obtain the estimates for according discrepancies, and therefore, it emerges the possibility to estimate the statistical properties of {z n }.

Notations.
N, Z, G denote, respectively, the sets of naturals, rational integers and Gaussian integers numbers, i.e.
p be the prime rational number congruent with 3 modulo 4; α, β, γ be the Gaussian integers; Z p m (respectively, Z * p m ) be denote the complete (respectively, reduced) residue system modulo p m over Z; G p m (respectively, G * p m ) be denote the complete (respectively, reduced) residue system modulo p m over G; ν p (α) be the nonnegative integer a such that α ≡ 0 (mod p a ), α ≡ 0 (mod p a+1 ).

Generator (3) we consider under conditions
These conditions ensure the sequence {z n } will be not stopped and will have the big enough period.We need the following lemmas.
Proof.If γ ≡ 0 (mod p m ), then the assertion is clear.Let ν p (γ) = a < m.Then for k = 1 from the congruence (here and elsewhere, all multiplicative inverses be set modulo p m ).Then Now, after m iterations, we obtain z ′′ ≡ z ′ (mod p m ), i.e., for k = 1 the assertion proved. Let Consider the congruence ), then we have Let assume that z ′ ≡ z ′′ (mod p m−νp(γ) ).But then Hence, after m νp(γ) steps we get contradiction with an assumption that z ′ ≡ z ′′ (mod p m−νp(γ) ), from where it follows the assertion of Lemma 1.
Henceforth, we will write z k instead of Φ k (z 0 ), where z 0 is an initial value of recursion (3).
) be the polynomial with coefficients over G, and let (C, p) = 1.Then for every A ∈ G we have Proof.This assertion is the corollary of the estimate of linear sum and of the analogue of estimate of the Gauss sum over finite field.
In order that to prove the relations (6), let consider the following matrices We have, where E = 1 0 0 1 , O = 0 0 0 0 . Moreover, Calculating z n+1 , z n+2 from expression for z n in (5) we get the relations where 1 ≤ ℓ ≤ r.

Simple calculations give
Now, from ( 7)-( 10) we obtain for even n modulo p m , integers z n generated by recursion (3), we have where , moreover, all multiplicative inverses take modulo p m .Proof.From ( 5) it is clear that for every n only one summand in the numerator and denominator is coprime with p. Thus, in formula for z n , multiplying the numerator and denominator of fraction by multiplicative inverse to denominator, and using p-adic factorization for and after simple calculations we obtain the representations for z 2n and z 2n+1 in powers of z 0 and z −1 0 modulo p m .

Main Results
For arbitrary Gaussian integers h 1 , h 2 let consider the sum .
Here, we consider z k , z ℓ as functions in z 0 with description in Lemma 4.
Proof.Let's begin with assumption that k and ℓ are the nonnegative integers with different parity.And let us agree to write z instead z 0 .By the Lemma 4 we may write where Then, we have z Hence, ).Now, by substituting the value of z from (15) and by summing over v, in virtue of classical estimation of the complete linear sum over G p , we deduce where F 1 (u, u −1 ) be the polynomial with coefficients from G p m−s−2 and has the same view as the polynomial F (u, u −1 ).Continuing these discourses, we obtain the assertion of lemma for k ≡ ℓ (mod 2).
Let k and ℓ be the numbers of same parity (for example, k = 2k 1 , ℓ = 2ℓ 1 ).Then modulo p m−s , we have where By induction on j it is easy to show that h (0) 1 ℓ ≡ 0 (mod p t ) for some t.Thus, as above, we conclude By virtue of lemma we infer In case k ≡ ℓ ≡ 1 (mod 2) the proof is the same.
For 1 ≤ N ≤ τ we define the sum Theorem 5. Let {z n } be the sequence generated by recursion (3) with a period τ = 2p m−ν , and let 2ν < µ.Then Proof.As τ = 2p m−ν is a maximal possible period for the sequences generated by (3), we may conclude that gcd(a − z 2 , p) = gcd(1 − az −2 , p) = 1.And, therefore, by Corollary 1 of Lemma 2 we obtain where the polynomials F (u) and G(u) have the following representation Therefore, the last two sums on the right in (20) may be simply reduced to complete linear sums, such that where δ = e This theorem is the corollary of Lemma 2 and estimate of sum through the compete sum.

Conclusion
In conclusion, we note that the estimates of exponential sums, obtained in Theorems 1-4, are essencially use the representation of z n in the form of polynomials in z 0 , z −1 0 or in n.These representations are also allowed to obtain the nontrivial estimates for exponential sums over s-dimentional points of the form (z n , z n+1 , . . ., z n+s−1 ), and so it makes possible to investigate such sums on passes the s-dimentional serial tests on statistical independence of the elements of sequence {z n }.