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%I #30 Feb 28 2024 12:44:31
%S 1,3,217,862,1177,1207,1219,3937,8743,9481,13822,18137,19567,19849,
%T 20057,20257,20299,20437,33607,57337,91847,96217,100579,103897,154969,
%U 157921,158623,228889,233047,304117,324817,325579,329057,330529,537817,595417,608287
%N Numbers n such that there is an integer k with the property that k^tau(n) = sigma(n).
%C Corresponding values of k: 1, 2, 4, 6, 6, 6, 6, 8, 10, 10, 12, 12, 12, 12, 12, 12, 12, 12, 14, 16, 18, 18, 18, 18, 20, 20, 20, 22, 22, 24, 24, 24, 24, 24, 28, 28, 28, ... (see A225369).
%C Conjecture: all terms are squarefree numbers.
%C Conjecture is false: p = (312^6 / 13) - 1 = 70955197267967 is prime, so sigma(9*p) = sigma(9)*sigma(p) = 13*(p+1) = 312^6 = 312^tau(9*p). - _Charlie Neder_, Oct 05 2018
%H Donovan Johnson, <a href="/A225239/b225239.txt">Table of n, a(n) for n = 1..500</a>
%e a(4) = 862 because sigma(862) = 1296 = 6^tau(862) = 6^4; k = 6.
%o (PARI) c=1; write("b225239.txt", c " " 1); for(n=2, 1943881801, s=sigma(n); if(ispower(s), nd=numdiv(n); r=round(sqrtn(s, nd)); if(r^nd==s, c++; write("b225239.txt", c " " n)))) /* _Donovan Johnson_, May 05 2013 */
%o (PARI) isok(n) = if (n==1, return(1)); my(s=sigma(n)); if(ispower(s), my(nd=numdiv(n)); r=sqrtnint(s, nd); (r^nd==s);); \\ _Michel Marcus_, Feb 19 2020
%Y Cf. A000005 (tau(n): number of divisors of n).
%Y Cf. A000203 (sigma(n): sum of divisors of n).
%Y Cf. A051281 (sigma(n) is a power of tau(n)), A225369.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, May 04 2013
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