A264154 For numbers m such that rad(n) divides sigma(n), this sequence gives the minimum exponent k such that sigma(m)^k divides m.
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, 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, 3, 3, 4, 3, 11, 4, 7, 3, 2, 2, 5, 3, 3, 5, 3, 4, 4, 7, 4
Offset: 1
Keywords
Examples
A175200(2) is 6, and for 6, sigma(6)^k/6 is already an integer with k=1, so a(2)=6. 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.
Links
- Michel Marcus, Table of n, a(n) for n = 1..10000
Programs
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PARI
fk(s, m) = {j = 1; while(denominator(s^j/m) != 1, j++); j;} rad(n) = factorback(factorint(n)[, 1]); lista(nn) = {for (n=1, nn, if (denominator(sigma(n)/rad(n)) == 1, k = fk(sigma(n), n); print1(k, ", ");););}