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A358557
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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).
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2
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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MATHEMATICA
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Select[Range[100], Denominator[HarmonicNumber[#]] < LCM @@ Range[#] &] (* Amiram Eldar, Nov 25 2022 *)
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PROG
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(PARI) isok(n) = lcm(vector(n, i, i)) <> denominator(sum(i=1, n, 1/i)); \\ Thomas Scheuerle, Nov 23 2022
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CROSSREFS
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Cf. A330680 (numbers that begin a run of consecutive integers not in the sequence).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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