A351315 Sum of the 9th powers of the square divisors of n.
1, 1, 1, 262145, 1, 1, 1, 262145, 387420490, 1, 1, 262145, 1, 1, 1, 68719738881, 1, 387420490, 1, 262145, 1, 1, 1, 262145, 3814697265626, 1, 387420490, 262145, 1, 1, 1, 68719738881, 1, 1, 1, 101560344351050, 1, 1, 1, 262145, 1, 1, 1, 262145, 387420490, 1, 1, 68719738881
Offset: 1
Examples
a(16) = 68719738881; a(16) = Sum_{d^2|16} (d^2)^9 = (1^2)^9 + (2^2)^9 + (4^2)^9 = 68719738881.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := (p^(18*(1 + Floor[e/2])) - 1)/(p^18 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 07 2022 *) snp[n_]:=Total[Select[Divisors[n],IntegerQ[Sqrt[#]]&]^9]; Array[snp,50] (* Harvey P. Dale, May 25 2025 *)
Formula
a(n) = Sum_{d^2|n} (d^2)^9.
Multiplicative with a(p) = (p^(18*(1+floor(e/2))) - 1)/(p^18 - 1). - Amiram Eldar, Feb 07 2022
From Amiram Eldar, Sep 20 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-18).
Sum_{k=1..n} a(k) ~ (zeta(19/2)/19) * n^(19/2). (End)
a(n) = Sum_{d|n} d^9 * c(d), where c = A010052. - Wesley Ivan Hurt, Jun 21 2024
a(n) = Sum_{d|n} lambda(d)*d^9*sigma_9(n/d), where lambda = A008836. - Ridouane Oudra, Jul 19 2025
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