A351439 Number of prime factors p of n such that p^(1+valuation(n,p)) divides sigma(n).

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Antti Karttunen, Feb 16 2022

Keywords

• nonn

Examples

			For n = 30 = 2*3*5, sigma(30) = 72 = 2^3 * 3^2 and thus for two of the three prime factors of 30, a higher power of the same prime divides sigma(30), and therefore a(30) = 2.
		

Links

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

Crossrefs

Cf. A000203, A001221, A007691, A286561, A351539.

Cf. also A336352.

Programs

Mathematica

{0}~Join~Table[Function[m, Count[FactorInteger[n][[All, 1]], ?(Mod[m, #^(1 + IntegerExponent[n, #])] == 0 &)]][DivisorSigma[1, n]], {n, 2, 108}] (* _Michael De Vlieger, Feb 16 2022 *)

PARI

A351439(n) = { my(f=factor(n),s=sigma(n)); sum(k=1,#f~,(0==(s%(f[k,1]^(1+f[k,2]))))); };

Formula

For all n >= 1, a(n) <= A001221(n). [Apparently this is sharp for n > 1].

For all n >= 1, a(n) >= A351539(n).