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A349258
a(n) is the number of prime powers (not including 1) that are infinitary divisors of n.
8
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
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
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Nov 12 2021
EXTENSIONS
Wrong comment removed by Amiram Eldar, Sep 22 2023
STATUS
approved