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

2,1

Comments

For prime p and m > 0, a(p^m) = p^(m+1). - Muniru A Asiru, Nov 23 2018

Links

David A. Corneth, Table of n, a(n) for n = 2..10001 (first 499 terms by Muniru A Asiru)

Formula

a(n) = ceiling(n^(1 + 1/(tau(n)-1))). - Jon E. Schoenfield, Nov 22 2018

Example

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.

Mathematica

Array[Block[{k = 1}, While[k^(#2 - 1) < #1^#2, k++] & @@ {#, DivisorSigma[0, #]}; k] &, 55, 2] (* Michael De Vlieger, Oct 10 2018 *)

Prog

(PARI) a(n) = my(nd = numdiv(n)); res = ceil(n ^ (nd / (nd - 1))); while(res^(nd-1) >= n^nd, res--); res+1
(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

Crossrefs

Cf. A000005, A291899.

Sequence in context: A063718 A063748 A121920 * A065642 A285109 A217579

Adjacent sequences: A318276 A318277 A318278 * A318280 A318281 A318282

Keyword

nonn,easy

Author

David A. Corneth, Oct 09 2018

Extensions

Correct value a(27) = 81 inserted by Muniru A Asiru, Nov 22 2018

Status

approved