

A273038


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


1



1, 4, 11, 33, 35, 11, 50, 67, 94, 35, 113, 33, 197, 50, 35, 249, 247, 94, 276, 35, 50, 113, 427, 67, 354, 197, 453, 50, 765, 35, 855, 544, 113, 247, 50, 94, 1130, 276, 197, 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
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OFFSET

1,2


COMMENTS

A proof of the existence of a(n) for all n was given by Vladimir Shevelev, May 14 2016, as follows:
(Start)
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."
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)
where p_i=prime(i).
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.
Let us show that, when N runs through A067128, k_1 is also unbounded.
Indeed, suppose k_1 is bounded. Consider a number N_1 with PPF N_1=2^(k_1+x)*...*p_(n1)^k_(n1) such that all powers p^i , i=2,...,n1, are the same as in (1) and satisfy 2^x<p_n^k_n. (2)
Then N_1<N. Let us try to choose x so that d(N_1)>d(N).
We want (k_1+x+1)*...*(k_(n1)+1)>(k_1+1)*...*(k_(n1)+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.
So, by (2), (k_1+1)*k_n<x<k_n*log_2(p_n). (3)
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.
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.
Suppose m=2^m_1*3^m_2*...*p_r^m_r.
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).
So there exists a large M_m such that for all N from A067128 > M_m, mN.
(End)


LINKS



CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



