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Sum of the cubes of the square divisors of n.
12

%I #27 Jun 20 2024 21:09:48

%S 1,1,1,65,1,1,1,65,730,1,1,65,1,1,1,4161,1,730,1,65,1,1,1,65,15626,1,

%T 730,65,1,1,1,4161,1,1,1,47450,1,1,1,65,1,1,1,65,730,1,1,4161,117650,

%U 15626,1,65,1,730,1,65,1,1,1,65,1,1,730,266305,1,1,1,65,1,1,1,47450,1

%N Sum of the cubes of the square divisors of n.

%C Inverse Möbius transform of n^3 * c(n), where c(n) is the characteristic function of squares (A010052). - _Wesley Ivan Hurt_, Jun 20 2024

%H Michael De Vlieger, <a href="/A351308/b351308.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d^2|n} (d^2)^3.

%F Multiplicative with a(p) = (p^(6*(1+floor(e/2))) - 1)/(p^6 - 1). - _Amiram Eldar_, Feb 07 2022

%F From _Amiram Eldar_, Sep 19 2023: (Start)

%F Dirichlet g.f.: zeta(s) * zeta(2*s-6).

%F Sum_{k=1..n} a(k) ~ (zeta(7/2)/7) * n^(7/2). (End)

%F G.f.: Sum_{k>=1} k^6 * x^(k^2) / (1 - x^(k^2)). - _Ilya Gutkovskiy_, Jun 05 2024

%F a(n) = Sum_{d|n} d^3 * c(d), where c = A010052. - _Wesley Ivan Hurt_, Jun 20 2024

%e a(16) = 4161; a(16) = Sum_{d^2|16} (d^2)^3 = (1^2)^3 + (2^2)^3 + (4^2)^3 = 4161.

%t f[p_, e_] := (p^(6*(1 + Floor[e/2])) - 1)/(p^6 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Feb 07 2022 *)

%o (PARI) a(n) = sumdiv(n, d, if (issquare(d), d^3)); \\ _Michel Marcus_, Mar 24 2023

%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), this sequence (k=3), A351309 (k=4), A351310 (k=5), A351311 (k=6), A351313 (k=7), A351314 (k=8), A351315 (k=9), A351315 (k=10).

%Y Cf. A010052, A261804 (zeta(7/2)).

%K nonn,easy,mult

%O 1,4

%A _Wesley Ivan Hurt_, Feb 06 2022