OFFSET
1,2
COMMENTS
Let f be an arbitrary arithmetic function. Define the sequence a(f; n) by a(f; n) = Sum_{i=1..n, k=1..n} f(n / gcd(gcd(i,k),n)) for n > 0. Then a(f; n) equals inverse Moebius transform of f(n) * A007434(n) for n > 0; if f is multiplicative then a(f; n) is multiplicative; this sequence uses f(n) = n (see formula section).
LINKS
Aloe Poliszuk, Table of n, a(n) for n = 1..10000
FORMULA
Multiplicative with a(p^e) = p^(3*e) - (p-1) * (p^(3*e) - 1) / (p^3 - 1) for prime p and e >= 0.
Dirichlet g.f.: Sum_{n>0} a(n) / n^s = zeta(s-3) * zeta(s) / zeta(s-1).
a(n) = Sum_{i=1..n, k=1..n} n / gcd(gcd(i,k),n) for n > 0.
Sum_{k=1..n} a(k) ~ c * n^4, where c = Pi^4/(360*zeta(3)) = 0.225098... . - Amiram Eldar, Oct 16 2022
a(n) = Sum_{d|n} phi(n/d) * (n/d)^2 * sigma_2(d^2)/sigma(d^2). - Seiichi Manyama, May 24 2024
a(n) = Sum_{1 <= x_1, x_2 <= n} ( gcd(x_1, n)/gcd(x_1, x_2, n) )^2. - Seiichi Manyama, May 25 2024
From Aloe Poliszuk, Nov 12 2025: (Start)
a(n) = Sum_{d|n} (n/d)^3 * A023900(d).
a(n) = Sum_{d|n} d * A007434(d). (End)
MATHEMATICA
f[p_, e_] := p^(3*e) - (p - 1)*(p^(3*e) - 1)/(p^3 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Jan 19 2022 *)
PROG
(Python)
from math import prod
from sympy import factorint
def A350156(n): return prod((q:=p**(3*e))-(p-1)*(q-1)//(p**3-1) for p, e in factorint(n).items()) # Chai Wah Wu, Mar 04 2025
(PARI) a(n) = sumdiv(n, d, d*moebius(d)*sigma(n/d, 3)); \\ Aloe Poliszuk, Nov 12 2025
CROSSREFS
KEYWORD
nonn,easy,mult
AUTHOR
Werner Schulte, Jan 19 2022
STATUS
approved
