A351539 Number of odd prime factors p of n such that p^(1+valuation(n,p)) divides sigma(n).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Feb 16 2022

Keywords

• nonn

Examples

			For n = 2235 = 3 * 5 * 149, sigma(2235) = 3600 = 2^4 * 3^2 * 5^2, and thus for two of the three odd prime factors of 2235, a higher power of the same prime divides sigma(2235), and therefore a(2235) = 2. This is also the least n for which a(n) > 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A000203, A001221, A286561, A351439, A351540 (positions of nonzero terms).

Programs

Mathematica

{0}~Join~Table[Function[m, Count[DeleteCases[FactorInteger[n][[All, 1]], 2], ?(Mod[m, #^(1 + IntegerExponent[n, #])] == 0 &)]][DivisorSigma[1, n]], {n, 2, 120}] (* _Michael De Vlieger, Feb 16 2022 *)

PARI

A351539(n) = { my(f=factor(n),s=sigma(n)); sum(k=1,#f~,(f[k,1]%2)&&(0==(s%(f[k,1]^(1+f[k,2]))))); };

Formula

For all n >= 1, a(n) <= A351439(n) <= A001221(n).