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Sum of the 4th powers of the square divisors of n.
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%I #28 Jun 29 2024 22:46:22

%S 1,1,1,257,1,1,1,257,6562,1,1,257,1,1,1,65793,1,6562,1,257,1,1,1,257,

%T 390626,1,6562,257,1,1,1,65793,1,1,1,1686434,1,1,1,257,1,1,1,257,6562,

%U 1,1,65793,5764802,390626,1,257,1,6562,1,257,1,1,1,257,1,1,6562,16843009,1

%N Sum of the 4th powers of the square divisors of n.

%C Inverse Möbius transform of n^4 * c(n), where c(n) is the characteristic function of squares (A010052). - _Wesley Ivan Hurt_, Jun 29 2024

%H Michael De Vlieger, <a href="/A351309/b351309.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d^2|n} (d^2)^4.

%F Multiplicative with a(p) = (p^(8*(1+floor(e/2))) - 1)/(p^8 - 1). - _Amiram Eldar_, Feb 07 2022

%F From _Amiram Eldar_, Sep 20 2023: (Start)

%F Dirichlet g.f.: zeta(s) * zeta(2*s-8).

%F Sum_{k=1..n} a(k) ~ (zeta(9/2)/9) * n^(9/2). (End)

%F G.f.: Sum_{k>=1} k^8 * x^(k^2) / (1 - x^(k^2)). - _Ilya Gutkovskiy_, Jun 05 2024

%F a(n) = Sum_{d|n} d^4 * c(d), where c = A010052. - _Wesley Ivan Hurt_, Jun 29 2024

%e a(16) = 65793; a(16) = Sum_{d^2|16} (d^2)^4 = (1^2)^4 + (2^2)^4 + (4^2)^4 = 65793.

%t 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 *)

%t Table[Total[Select[Divisors[n],IntegerQ[Sqrt[#]]&]^4],{n,70}] (* _Harvey P. Dale_, Feb 11 2023 *)

%o (PARI) a(n) = sumdiv(n, d, if (issquare(d), d^4)); \\ _Michel Marcus_, Jun 05 2024

%Y 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).

%Y Cf. A010052.

%K nonn,easy,mult

%O 1,4

%A _Wesley Ivan Hurt_, Feb 06 2022