A369210 Numbers k such that the number of divisors of k^2 is a power of 3.
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 48, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
pow3q[n_] := n == 3^IntegerExponent[n, 3]; Select[Range[100], pow3q[DivisorSigma[0, #^2]] &]
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PARI
ispow3(n) = n == 3^valuation(n, 3); is(n) = ispow3(numdiv(n^2));
Formula
Sum_{n>=1} 1/a(n)^2 = Product_{p prime} Sum_{k>=0} 1/p^(3^k-1) = 1.52478035628964060288... .
Comments