A337050 Numbers without an exponent 2 in their prime factorization.
1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87
Offset: 1
Examples
6 = 2^1 * 3^1 is a term since none of the exponents in its prime factorization is equal to 2. 9 = 3^2 is not a term since it has an exponent 2 in its prime factorization.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Ertan Elma and Greg Martin, Distribution of the number of prime factors with a given multiplicity, Canadian Mathematical Bulletin, Vol. 67, No. 4 (2024), pp. 1107-1122; arXiv preprint, arXiv:2406.04574 [math.NT], 2024.
- D. Suryanarayana, Semi-k-free integers, Elemente der Mathematik, Vol. 26 (1971), pp. 39-40.
- D. Suryanarayana and R. Sitaramachandra Rao, Distribution of semi-k-free integers, Proceedings of the American Mathematical Society, Vol. 37, No. 2 (1973), pp. 340-346.
Crossrefs
Programs
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Maple
q:= n-> andmap(i-> i[2]<>2, ifactors(n)[2]): select(q, [$1..100])[]; # Alois P. Heinz, Aug 12 2020
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Mathematica
Select[Range[100], !MemberQ[FactorInteger[#][[;;, 2]], 2] &]
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PARI
is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 2] == 2, return(0))); 1; } \\ Amiram Eldar, Oct 21 2023
Formula
Sum_{n>=1} 1/a(n)^s = zeta(s) * Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s)), for s > 1. - Amiram Eldar, Oct 21 2023
Comments