cp's OEIS Frontend

A273038 Least k such that for all m >= k, A067128(m) is divisible by n.

%I A273038 #27 Dec 23 2024 14:53:44
%S A273038 1,4,11,33,35,11,50,67,94,35,113,33,197,50,35,249,247,94,276,35,50,
%T A273038 113,427,67,354,197,453,50,765,35,855,544,113,247,50,94,1130,276,197,
%U A273038 67,1274,50,1457,113,94,427,1853,249,994,354,247,197,3433,453,113,67,276,765,3757,35,4123,855,94,1703,197,113,4465
%N A273038 Least k such that for all m >= k, A067128(m) is divisible by n.
%C A273038 A proof of the existence of a(n) for all n was given by _Vladimir Shevelev_, May 14 2016, as follows:
%C A273038 (Start)
%C A273038 I give a proof of the existence of k in new David's sequence A273038: "Least k such that for all m >= k, A067128(m) is divisible by n."
%C A273038 Let us change the notation. Suppose N in A067128 has prime power factorization (PPF) N=2^k_1*...*p_n^k_n, k_n>=1,     (1)
%C A273038 where p_i=prime(i).
%C A273038 From my theorem in A273015 it follows that, when N runs through A067128, p_n in (1) is unbounded and, moreover, tends to infinity, when N tends to infinity.
%C A273038 Let us show that, when N runs through A067128, k_1 is also unbounded.
%C A273038 Indeed, suppose k_1 is bounded. Consider a number N_1 with PPF N_1=2^(k_1+x)*...*p_(n-1)^k_(n-1) such that all powers p^i , i=2,...,n-1, are the same as in (1) and satisfy 2^x<p_n^k_n. (2)
%C A273038 Then N_1<N.  Let us try to choose x so that d(N_1)>d(N).
%C A273038 We want (k_1+x+1)*...*(k_(n-1)+1)>(k_1+1)*...*(k_(n-1)+1)*(k_n+1), or k_1+x+1>(k_1+1)*(k_n+1)=k_1*k_n+k_n+k_1+1, or x>(k_1+1)*k_n.
%C A273038 So, by (2),  (k_1+1)*k_n<x<k_n*log_2(p_n).  (3)
%C A273038 Since by hypothesis k_1 is bounded, for large n we can choose the required x, which gives a contradiction. So k_1 is unbounded.
%C A273038 Moreover, we see that k_1 tends to infinity as log_2(p_n), n=n(N), when N tends to infinity, otherwise (3) again leads to contradiction.
%C A273038 Suppose m=2^m_1*3^m_2*...*p_r^m_r.
%C A273038 We can choose k_1 > m_1. In the same way we prove that k_2 tends to infinity and choose k_2 > m_2,..., and so on. k_r tends to infinity and we choose k_r > m_r.  All k_i , i=1,...,r tend to infinity at least as log_p_r(p_n), n=n(N).
%C A273038 So there exists a large M_m such that for all N from A067128 > M_m, m|N.
%C A273038 (End)
%H A273038 Vladimir Shevelev, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-May/016399.html">Posting to Sequence Fans Mailing List</a>, May 14 2016.
%Y A273038 Cf. A034287, A067128, A273014, A273015, A273016, A273018.
%K A273038 nonn
%O A273038 1,2
%A A273038 _David A. Corneth_, May 13 2016