A361430 Multiplicative with a(p^e) = e - 1.

1, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 4, 0, 0, 0, 1, 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, 5, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Vaclav Kotesovec, Mar 11 2023

Keywords

• nonn
• easy
• mult

Comments

a(n) is the number of coreful divisors d of n such that n/d is also a coreful divisor of n (a coreful divisor d of a number n is a divisor with the same set of distinct prime factors as n, see A307958). - Amiram Eldar, Aug 15 2023

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Vaclav Kotesovec, Graph - the asymptotic ratio (10^9 terms)

Crossrefs

Cf. A001694, A298826, A307958, A335850 (indices of records).

Cf. A005361 (multiplicative with a(p^e) = e), A000005 (e+1), A343443 (e+2), A360997 (e+3), A360908 (2*e-1), A360910 (3*e-1), A360911 (3*e-2).

Programs

Mathematica

g[p_, e_] := e-1; a[1] = 1; a[n_] := Times @@ g @@@ FactorInteger[n]; Array[a, 200]

PARI

for(n=1, 100, print1(direuler(p=2, n, 1 + 1/(1 - 1/X)^2)[n], ", "))

PARI

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = f[k,2]-1; f[k,2] = 1); factorback(f); \\ Michel Marcus, Mar 13 2023

Python

from math import prod
from sympy import factorint
def A361430(n): return prod(e-1 for e in factorint(n).values()) # Chai Wah Wu, Apr 15 2025

Formula

Dirichlet g.f.: Product_{primes p} (1 + 1/(p^s - 1)^2).

Dirichlet g.f.: zeta(2*s) * zeta(3*s)^2 * Product_{primes p} (1 + 2/p^(4*s) + 2/p^(5*s) - 1/p^(6*s) - 2/p^(7*s) - 2/p^(8*s)).

Let f(s) = Product_{primes p} (1 + 2/p^(4*s) + 2/p^(5*s) - 1/p^(6*s) - 2/p^(7*s) - 2/p^(8*s)), then

Sum_{k=1..n} a(k) ~ f(1/2) * zeta(3/2)^2 * sqrt(n) + zeta(2/3) * (f(1/3) * (log(n) + 6*gamma - 3 + 2*zeta'(2/3)/zeta(2/3)) + f'(1/3)) * n^(1/3) / 3, where

f(1/2) = Product_{primes p} (1 + 2/p^2 + 2/p^(5/2) - 1/p^3 - 2/p^(7/2) - 2/p^4) = 2.20286226691565931157047065666916419062717171359087693723221239...

f(1/3) = Product_{primes p} (1 + 2/p^(4/3) + 2/p^(5/3) - 1/p^2 - 2/p^(7/3) - 2/p^(8/3)) = 6.250573144372477079986352917664218040797528021629950408099536...

f'(1/3) = f(1/3) * Sum_{primes p} (-2*(-8 + p^(1/3) + 4*p^(2/3)) * log(p) / (-2 + p^(2/3) + p - p^(5/3) + p^2)) = -90.898558294301467740374653006294640945295... and gamma is the Euler-Mascheroni constant A001620.

Conjecture: a(n) = abs(A298826(n)).

a(n) > 0 if and only if n is powerful (A001694). - Amiram Eldar, Aug 15 2023