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%I #12 Dec 19 2017 14:11:30
%S 1,2,4,4,8,8,8,8,16,16,16,16,16,16,16,16,32,32,32,32,32,32,32,32,32,
%T 32,32,32,32,32,32,32,64,64,64,64,64,64,64,64,64,64,64,64,64,64,64,64,
%U 64,64,64,64,64,64,64,64,64,64,64,64,64,64,64,64,128,128
%N Smallest k such that either k >= n and k is a power of 2, or k >= 5n/3 and the prime divisors of k are precisely 2 and 5.
%C First disagreement with A062383(n-1) is at n = 129.
%C For n > 2, a(n) is not squarefree. - _Iain Fox_, Dec 17 2017
%H Eric M. Schmidt, <a href="/A296613/b296613.txt">Table of n, a(n) for n = 1..10000</a>
%H Bernadette Faye, Florian Luca, Pieter Moree, <a href="https://arxiv.org/abs/1708.03563">On the discriminator of Lucas sequences</a>, arXiv:1708.03563 [math.NT], 2017, Theorem 1.
%o (PARI) a(n) = for(k=n, +oo, if(k == 2^valuation(k, 2) || (k >= 5*n/3 && factor(k)[, 1] == [2, 5]~), return(k))) \\ _Iain Fox_, Dec 17 2017
%Y Cf. A033846.
%K nonn
%O 1,2
%A _Eric M. Schmidt_, Dec 16 2017
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