A358557 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).

6, 7, 8, 18, 19, 20, 21, 22, 23, 24, 25, 26, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 100, 101, 102

Offset: 1

Author

Yifan Xie, Nov 22 2022

Keywords

• nonn

Comments

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).

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).

The sequence is the answer to Problem 23 of the 2022 AMC12A.

Links

• Yifan Xie, Table of n, a(n) for n = 1..10000
• Art of Problem Solving Website, The Mathematical Association of America, Entry 2022 AMC12/AHSME

Crossrefs

Cf. A001008/A002805 (harmonic numbers), A003418 (LCM).

Cf. A110566 (common factor).

Cf. A098464 (complement), A112813.

Cf. A330680 (numbers that begin a run of consecutive integers not in the sequence).

Programs

Mathematica

Select[Range[100], Denominator[HarmonicNumber[#]] < LCM @@ Range[#] &] (* Amiram Eldar, Nov 25 2022 *)

PARI

isok(n) = lcm(vector(n, i, i)) <> denominator(sum(i=1, n, 1/i)); \\ Thomas Scheuerle, Nov 23 2022

Formula

A110566(a(n)) > 1. - Thomas Scheuerle, Nov 23 2022