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A321559
a(n) = Sum_{d divides n} (-1)^(d + n/d) * d^3.
3
1, -9, 28, -57, 126, -252, 344, -441, 757, -1134, 1332, -1596, 2198, -3096, 3528, -3513, 4914, -6813, 6860, -7182, 9632, -11988, 12168, -12348, 15751, -19782, 20440, -19608, 24390, -31752, 29792, -28089, 37296, -44226, 43344, -43149, 50654
OFFSET
1,2
FORMULA
G.f.: Sum_{k>=1} (-1)^(k+1)*k^3*x^k/(1 + x^k). - Ilya Gutkovskiy, Nov 27 2018
From Peter Bala, Jan 29 2022: (Start)
Multiplicative with a(2^k) = - 3*(2^(3*k+1) + 5)/7 for k >= 1 and a(p^k) = (p^(3*k+3) - 1)/(p^3 - 1) for odd prime p.
n^3 = (-1)^(n+1)*Sum_{d divides n} A067856(n/d)*a(d). (End)
MATHEMATICA
a[n_] := DivisorSum[n, (-1)^(# + n/#)*#^3 &]; Array[a, 50] (* Amiram Eldar, Nov 27 2018 *)
PROG
(PARI) apply( A321559(n)=sumdiv(n, d, (-1)^(n\d-d)*d^3), [1..30]) \\ M. F. Hasler, Nov 26 2018
(Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (&+[(-1)^(k+1)*k^3*x^k/(1 + x^k) : k in [1..2*m]]) )); // G. C. Greubel, Nov 28 2018
(Sage) s=(sum((-1)^(k+1)*k^3*x^k/(1 + x^k) for k in (1..50))).series(x, 50); a = s.coefficients(x, sparse=False); a[1:] # G. C. Greubel, Nov 28 2018
CROSSREFS
Column k=3 of A322083.
Cf. A321543 - A321565, A321807 - A321836 for similar sequences.
Sequence in context: A063155 A366863 A135705 * A041359 A034126 A034677
KEYWORD
sign,mult
AUTHOR
N. J. A. Sloane, Nov 23 2018
STATUS
approved