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

1,30

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to sigma(n)

Formula

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

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

Example

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.

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 *)

Prog

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

Crossrefs

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

Cf. also A336352.

Sequence in context: A323913 A014082 A322583 * A102354 A370222 A349615

Adjacent sequences: A351436 A351437 A351438 * A351440 A351441 A351442

Keyword

nonn

Author

Antti Karttunen, Feb 16 2022

Status

approved