A349258 a(n) is the number of prime powers (not including 1) that are infinitary divisors of n.

0, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 3, 2, 1, 3, 1, 3, 2, 2, 2, 2, 1, 2, 2, 4, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 4, 2, 4, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 1, 4, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 4, 1, 3, 2, 2, 2, 2, 2, 4, 1, 2, 2, 2, 1, 3, 1, 4, 3

Offset

1,6

Comments

The total number of prime powers (not including 1) that divide n is A001222(n).

For each n, all the prime powers that are infinitary divisors of n are "Fermi-Dirac primes" (A050376).

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

Additive with a(p^e) = 2^A000120(e) - 1.

a(n) <= A001222(n), with equality if and only if n is in A036537.

a(n) <= A037445(n) - 1, with equality if and only if n is a prime power (including 1, A000961).

Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 0.28135949730844648114..., where f(x) = -(x+1) + (1-x) * Product_{k>=0} (1 + 2*x^(2^k)). - Amiram Eldar, Sep 29 2023

Example

12 has 4 infinitary divisors, 1, 3, 4 and 12. Two of these divisors, 3 and 4 = 2^2 are prime powers. Therefore a(12) = 2.

Mathematica

f[p_, e_] := 2^DigitCount[e, 2, 1] - 1; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

(PARI) A349258(n) = if(1==n, 0, vecsum(apply(x->(2^hammingweight(x))-1, factor(n)[, 2]))); \\ Antti Karttunen, Nov 12 2021

Crossrefs

Cf. A000120, A000961, A001222, A036537, A037445, A050376, A077609, A077761, A246655.

Sequence in context: A327518 A174532 A089242 * A349326 A185894 A214180

Adjacent sequences: A349255 A349256 A349257 * A349259 A349260 A349261

Keyword

nonn,easy

Author

Amiram Eldar, Nov 12 2021

Extensions

Wrong comment removed by Amiram Eldar, Sep 22 2023

Status

approved