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%I #12 Feb 14 2019 05:10:58
%S 1,1,2,1,3,3,3,1,3,3,2,2,2,3,4,5,1,3,2,2,7,1,2,4,3,3,2,3,3,2,5,2,2,3,
%T 3,5,2,3,7,3,3,3,5,3,4,2,5,3,2,7,2,3,3,3,5,2,7,2,6,2,5,3,2,3,3,2,3,1,
%U 3,3,4,3,11,4,7,3,2,2,5,3,3,5,3,4,4,7,4
%N For numbers m such that rad(n) divides sigma(n), this sequence gives the minimum exponent k such that sigma(m)^k divides m.
%H Michel Marcus, <a href="/A264154/b264154.txt">Table of n, a(n) for n = 1..10000</a>
%e A175200(2) is 6, and for 6, sigma(6)^k/6 is already an integer with k=1, so a(2)=6.
%e A175200(3) is 24, and for 24, sigma(24)/24 is not an integer while sigma(24)^2/24 is an integer, so a(3)=2.
%o (PARI) fk(s, m) = {j = 1; while(denominator(s^j/m) != 1, j++); j;}
%o rad(n) = factorback(factorint(n)[, 1]);
%o lista(nn) = {for (n=1, nn, if (denominator(sigma(n)/rad(n)) == 1, k = fk(sigma(n), n); print1(k, ", ");););}
%Y Cf. A000203 (sigma(n)), A007947 (rad(n)), A175200.
%K nonn
%O 1,3
%A _Michel Marcus_, Nov 06 2015
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