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%I #58 Jan 01 2024 08:02:53
%S 6,7,8,18,19,20,21,22,23,24,25,26,33,34,35,36,37,38,39,40,41,42,43,44,
%T 45,46,47,48,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,
%U 73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,100,101,102
%N Numbers k for which denominator(H(k)) < LCM(1..k), where harmonic numbers H(k) = Sum_{i=1..k} 1/i = r(k)/q(k).
%C LCM(1..k) is a common denominator for the harmonic numbers, and the present terms k are where the sum reduces to a smaller denominator (A002805).
%C We can find a prime p and a pair of positive integers t < p and o for every positive integer k that p^o*t <= k < p^o*(t+1). For positive integers i that are not divisible by p^o, a multiple of p will be added to the numerator of the reciprocal sum; for i's that are divisible by p^o, the number that will be added to the numerator of the reciprocal sum is divisible by r(t). So k is in the sequence if and only if p^o*t <= k < p^o*(t+1) where p is a prime and p divides r(t).
%C The sequence is the answer to Problem 23 of the 2022 AMC12A.
%H Yifan Xie, <a href="/A358557/b358557.txt">Table of n, a(n) for n = 1..10000</a>
%H Art of Problem Solving Website, The Mathematical Association of America, <a href="https://artofproblemsolving.com/community/c3200154_2022_amc_12ahsme">Entry 2022 AMC12/AHSME</a>
%F A110566(a(n)) > 1. - _Thomas Scheuerle_, Nov 23 2022
%t Select[Range[100], Denominator[HarmonicNumber[#]] < LCM @@ Range[#] &] (* _Amiram Eldar_, Nov 25 2022 *)
%o (PARI) isok(n) = lcm(vector(n, i, i)) <> denominator(sum(i=1, n, 1/i)); \\ _Thomas Scheuerle_, Nov 23 2022
%Y Cf. A001008/A002805 (harmonic numbers), A003418 (LCM).
%Y Cf. A110566 (common factor).
%Y Cf. A098464 (complement), A112813.
%Y Cf. A330680 (numbers that begin a run of consecutive integers not in the sequence).
%K nonn
%O 1,1
%A _Yifan Xie_, Nov 22 2022
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