A351309 Sum of the 4th powers of the square divisors of n.
1, 1, 1, 257, 1, 1, 1, 257, 6562, 1, 1, 257, 1, 1, 1, 65793, 1, 6562, 1, 257, 1, 1, 1, 257, 390626, 1, 6562, 257, 1, 1, 1, 65793, 1, 1, 1, 1686434, 1, 1, 1, 257, 1, 1, 1, 257, 6562, 1, 1, 65793, 5764802, 390626, 1, 257, 1, 6562, 1, 257, 1, 1, 1, 257, 1, 1, 6562, 16843009, 1
Offset: 1
Examples
a(16) = 65793; a(16) = Sum_{d^2|16} (d^2)^4 = (1^2)^4 + (2^2)^4 + (4^2)^4 = 65793.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[p_, e_] := (p^(8*(1 + Floor[e/2])) - 1)/(p^8 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 07 2022 *) Table[Total[Select[Divisors[n],IntegerQ[Sqrt[#]]&]^4],{n,70}] (* Harvey P. Dale, Feb 11 2023 *)
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PARI
a(n) = sumdiv(n, d, if (issquare(d), d^4)); \\ Michel Marcus, Jun 05 2024
Formula
a(n) = Sum_{d^2|n} (d^2)^4.
Multiplicative with a(p) = (p^(8*(1+floor(e/2))) - 1)/(p^8 - 1). - Amiram Eldar, Feb 07 2022
From Amiram Eldar, Sep 20 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-8).
Sum_{k=1..n} a(k) ~ (zeta(9/2)/9) * n^(9/2). (End)
G.f.: Sum_{k>=1} k^8 * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Jun 05 2024
a(n) = Sum_{d|n} d^4 * c(d), where c = A010052. - Wesley Ivan Hurt, Jun 29 2024
a(n) = Sum_{d|n} lambda(d)*d^4*sigma_4(n/d), where lambda = A008836. - Ridouane Oudra, Jul 19 2025
Comments