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A351309
Sum of the 4th powers of the square divisors of n.
11
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
1,4
COMMENTS
Inverse Möbius transform of n^4 * c(n), where c(n) is the characteristic function of squares (A010052). - Wesley Ivan Hurt, Jun 29 2024
LINKS
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
EXAMPLE
a(16) = 65793; a(16) = Sum_{d^2|16} (d^2)^4 = (1^2)^4 + (2^2)^4 + (4^2)^4 = 65793.
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 *)
PROG
(PARI) a(n) = sumdiv(n, d, if (issquare(d), d^4)); \\ Michel Marcus, Jun 05 2024
CROSSREFS
Sum of the k-th powers of the square divisors of n for k=0..10: A046951 (k=0), A035316 (k=1), A351307 (k=2), A351308 (k=3), this sequence (k=4), A351310 (k=5), A351311 (k=6), A351313 (k=7), A351314 (k=8), A351315 (k=9), A351315 (k=10).
Cf. A010052.
Sequence in context: A180699 A195529 A004217 * A051333 A273775 A182912
KEYWORD
nonn,easy,mult
AUTHOR
Wesley Ivan Hurt, Feb 06 2022
STATUS
approved