Uniform distribution of linear recurring sequences modulo prime powers.
If \(a_0,\dots, a_{d-1}\in \mathbb{Z}\) and \(u= \{u_n\}^\infty_{n=0}\) is a sequence satisfying the recurrence relation \(u_{n+d}= a_{d-1} u_{n+d-1}+\cdots+ a_0 u_n\) for \(n= 0,1,\dots\), then \(u\) is called a linear recurring sequence with defining coefficients \(a_0,\dots, a_{d-1}\) and initial values \(u_0,\dots, u_{d-1}\). The integer \(d\) is the order of the recurrence and \(P(x)= x^d- a_{d-1} x^{d-1}-\cdots- a_0\) is the characteristic polynomial. The author proves the following: Let \(p\in\mathbb{Z}\) be a prime, \(d\geq 2\) an integer, \(u\) a \(d\)th-order integer linear recurring sequence and \(S= (3d^2+ 9d)/2+ 1\). If \(u\) is uniformly distributed \(\text{mod\,}p^s\), then it is also uniformly distributed \(\text{mod\,} p^s\) for any \(s\in\mathbb{N}\). This answers a long-standing open question.
scientific article; zbMATH DE number 2029206
scientific article