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A352031
Sum of the cubes of the odd proper divisors of n.
11
0, 1, 1, 1, 1, 28, 1, 1, 28, 126, 1, 28, 1, 344, 153, 1, 1, 757, 1, 126, 371, 1332, 1, 28, 126, 2198, 757, 344, 1, 3528, 1, 1, 1359, 4914, 469, 757, 1, 6860, 2225, 126, 1, 9632, 1, 1332, 4257, 12168, 1, 28, 344, 15751, 4941, 2198, 1, 20440, 1457, 344, 6887, 24390, 1, 3528, 1
OFFSET
1,6
FORMULA
a(n) = Sum_{d|n, d<n, d odd} d^3.
G.f.: Sum_{k>=1} (2*k-1)^3 * x^(4*k-2) / (1 - x^(2*k-1)). - Ilya Gutkovskiy, Mar 02 2022
From Amiram Eldar, Oct 11 2023: (Start)
a(n) = A051000(n) - n^3*A000035(n).
Sum_{k=1..n} a(k) ~ c * n^4, where c = (zeta(4)-1)/8 = 0.0102904042... . (End)
EXAMPLE
a(10) = 126; a(10) = Sum_{d|10, d<10, d odd} d^3 = 1^3 + 5^3 = 126.
MATHEMATICA
f[2, e_] := 1; f[p_, e_] := (p^(3*e+3) - 1)/(p^3 - 1); a[1] = 0; a[n_] := Times @@ f @@@ FactorInteger[n] - If[OddQ[n], n^3, 0]; Array[a, 60] (* Amiram Eldar, Oct 11 2023 *)
PROG
(PARI) a(n) = sumdiv(n/2^valuation(n, 2), d, if ((d<n), d^3)); \\ Michel Marcus, Mar 02 2022
CROSSREFS
Sum of the k-th powers of the odd proper divisors of n for k=0..10: A091954 (k=0), A091570 (k=1), A351647 (k=2), this sequence (k=3), A352032 (k=4), A352033 (k=5), A352034 (k=6), A352035 (k=7), A352036 (k=8), A352037 (k=9), A352038 (k=10).
Sequence in context: A040784 A347162 A347174 * A174188 A176641 A040796
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Mar 01 2022
STATUS
approved