On the prime power factorization of \(n!\)
Let \(1=p_1<p_2<\dots\) be the sequence of primes. Let us denote by \(e_i(m)\) the integers satisfying \(p_{i}^{e_i(m)}\mid m\) and \(p_{i}^{e_i(m)+1}\nmid m\) (\(m\) is natural number). Let \(H_i(n)=e_i(n!).\) Motivated by a question of Erdős and Graham, \textit{D. Berend} showed, among other results, [J. Number Theory 64, 13-19 (1997; Zbl 0874.11025)] that for every \(k\geq 1\) there exists an infinite sequence of integers \(n_1<n_2<\dots \) with bounded gaps for which \(H_i(n_m)\equiv 0\pmod 2\), \(i=1,2,\dots ,k.\) In the present paper the authors generalize a result of Berend. They define the sequence \[ S(n)=H_i(u_jn+v_j)\pmod d_{1\leq i\leq k; 1\leq j\leq t}, \] where \(0<u_i\leq u; \;0\leq v_j\leq v\). (For \(v_j=0, u_j=j\) the case was investigated by Berend). For integers \(1\leq n<m\) if \(S(n)=S(m)\) then \(m\) is said to be a successor of \(n\) and \(n\) is a predecessor of \(m\). The authors prove that for every natural number \(n\) there exists at least one successor of \(n\) (and therefore infinitely many). This sequence has bounded gaps and this bound depends only on \(u,v,k,t,p_1,\dots ,p_k.\) As a consequence they prove given a positive \(k\) and \(\varepsilon_1,\varepsilon_2,\dots \varepsilon_k\in \{0,1\}\), if there exists a natural number \(n\) for which \(H_i(N)\equiv \varepsilon_i \pmod 2\), \(i=1,2,\dots ,k,\) then every block of \(m\) consecutive natural numbers contains an integer with the same property where \(m=m(p_1,p_2,\dots p_k).\)
scientific article; zbMATH DE number 1538250
scientific article