A352055 Sum of the 9th powers of the divisor complements of the odd proper divisors of n.

0, 512, 19683, 262144, 1953125, 10078208, 40353607, 134217728, 387440172, 1000000512, 2357947691, 5160042496, 10604499373, 20661047296, 38445332183, 68719476736, 118587876497, 198369368576, 322687697779, 512000262144, 794320419871, 1207269218304, 1801152661463, 2641941757952

Offset: 1

Author

Wesley Ivan Hurt, Mar 01 2022

Keywords

• nonn
• easy

Examples

			a(10) = 10^9 * Sum_{d|10, d<10, d odd} 1 / d^9 = 10^9 * (1/1^9 + 1/5^9) = 1000000512.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10000

Crossrefs

Sum of the k-th powers of the divisor complements of the odd proper divisors of n for k=0..10: A091954 (k=0), A352047 (k=1), A352048 (k=2), A352049 (k=3), A352050 (k=4), A352051 (k=5), A352052 (k=6), A352053 (k=7), A352054 (k=8), this sequence (k=9), A352056 (k=10).

Cf. A000035, A006519, A013668, A321813.

Programs

Mathematica

A352055[n_]:=DivisorSum[n,1/#^9&,#A352055,50] (* Paolo Xausa, Aug 10 2023 *)
a[n_] := DivisorSigma[-9, n/2^IntegerExponent[n, 2]] * n^9 - Mod[n, 2]; Array[a, 100] (* Amiram Eldar, Oct 13 2023 *)

PARI

a(n) = n^9 * sigma(n >> valuation(n, 2), -9) - n % 2; \\ Amiram Eldar, Oct 13 2023

Formula

a(n) = n^9 * Sum_{d|n, d

G.f.: Sum_{k>=2} k^9 * x^k / (1 - x^(2*k)). - Ilya Gutkovskiy, May 19 2023

From Amiram Eldar, Oct 13 2023: (Start)

a(n) = A321813(n) * A006519(n)^9 - A000035(n).

Sum_{k=1..n} a(k) = c * n^10 / 10, where c = 1023*zeta(10)/1024 = 1.0000170413... . (End)

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