|
%I #38 Feb 25 2024 06:00:24
%S 4,9,8,25,11,49,16,27,22,121,20,169,34,37,32,289,33,361,37,58,62,529,
%T 38,125,78,81,55,841,49,961,64,106,111,115,57,1369,128,133,68,1681,72,
%U 1849,94,97,165,2209,74,343,110,190,115,2809,96,210,100,220,225
%N a(n) is the least k such that k^(tau(n)-1) >= n^tau(n).
%C For prime p and m > 0, a(p^m) = p^(m+1). - _Muniru A Asiru_, Nov 23 2018
%H David A. Corneth, <a href="/A318279/b318279.txt">Table of n, a(n) for n = 2..10001</a> (first 499 terms by Muniru A Asiru)
%F a(n) = ceiling(n^(1 + 1/(tau(n)-1))). - _Jon E. Schoenfield_, Nov 22 2018
%e 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.
%t Array[Block[{k = 1}, While[k^(#2 - 1) < #1^#2, k++] & @@ {#, DivisorSigma[0, #]}; k] &, 55, 2] (* _Michael De Vlieger_, Oct 10 2018 *)
%o (PARI) a(n) = my(nd = numdiv(n)); res = ceil(n ^ (nd / (nd - 1))); while(res^(nd-1) >= n^nd, res--); res+1
%o (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
%Y Cf. A000005, A291899.
%K nonn,easy
%O 2,1
%A _David A. Corneth_, Oct 09 2018
%E Correct value a(27) = 81 inserted by _Muniru A Asiru_, Nov 22 2018
|
