A318279 a(n) is the least k such that k^(tau(n)-1) >= n^tau(n).
4, 9, 8, 25, 11, 49, 16, 27, 22, 121, 20, 169, 34, 37, 32, 289, 33, 361, 37, 58, 62, 529, 38, 125, 78, 81, 55, 841, 49, 961, 64, 106, 111, 115, 57, 1369, 128, 133, 68, 1681, 72, 1849, 94, 97, 165, 2209, 74, 343, 110, 190, 115, 2809, 96, 210, 100, 220, 225
Offset: 2
Examples
As tau(4) = 3, we look for the least k such that k^(3-1) >= 4^3, for which we find k = 8. Therefore, a(4) = 8.
Links
- David A. Corneth, Table of n, a(n) for n = 2..10001 (first 499 terms by Muniru A Asiru)
Programs
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GAP
List(List([2..57],n->Filtered([2..3000],k->k^(Tau(n)-1) >= n^Tau(n))),i->i[1]); # Muniru A Asiru, Oct 09 2018
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Mathematica
Array[Block[{k = 1}, While[k^(#2 - 1) < #1^#2, k++] & @@ {#, DivisorSigma[0, #]}; k] &, 55, 2] (* Michael De Vlieger, Oct 10 2018 *) -
PARI
a(n) = my(nd = numdiv(n)); res = ceil(n ^ (nd / (nd - 1))); while(res^(nd-1) >= n^nd, res--); res+1
Formula
a(n) = ceiling(n^(1 + 1/(tau(n)-1))). - Jon E. Schoenfield, Nov 22 2018
Extensions
Correct value a(27) = 81 inserted by Muniru A Asiru, Nov 22 2018
Comments