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a(n) = Sum_{d divides n} (-1)^(d + n/d) * d^3.
3

%I #24 Sep 08 2022 08:46:23

%S 1,-9,28,-57,126,-252,344,-441,757,-1134,1332,-1596,2198,-3096,3528,

%T -3513,4914,-6813,6860,-7182,9632,-11988,12168,-12348,15751,-19782,

%U 20440,-19608,24390,-31752,29792,-28089,37296,-44226,43344,-43149,50654

%N a(n) = Sum_{d divides n} (-1)^(d + n/d) * d^3.

%H G. C. Greubel, <a href="/A321559/b321559.txt">Table of n, a(n) for n = 1..1000</a>

%H Peter Bala, <a href="/A067856/a067856_1.pdf">A signed Dirichlet product of arithmetical functions</a>

%H J. W. L. Glaisher, <a href="https://books.google.com/books?id=bLs9AQAAMAAJ&amp;pg=RA1-PA1">On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares</a>, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).

%H <a href="/index/Ge#Glaisher">Index entries for sequences mentioned by Glaisher</a>

%F G.f.: Sum_{k>=1} (-1)^(k+1)*k^3*x^k/(1 + x^k). - _Ilya Gutkovskiy_, Nov 27 2018

%F From _Peter Bala_, Jan 29 2022: (Start)

%F 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.

%F n^3 = (-1)^(n+1)*Sum_{d divides n} A067856(n/d)*a(d). (End)

%t a[n_] := DivisorSum[n, (-1)^(# + n/#)*#^3 &]; Array[a, 50] (* _Amiram Eldar_, Nov 27 2018 *)

%o (PARI) apply( A321559(n)=sumdiv(n, d, (-1)^(n\d-d)*d^3), [1..30]) \\ _M. F. Hasler_, Nov 26 2018

%o (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

%o (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

%Y Column k=3 of A322083.

%Y Cf. A321543 - A321565, A321807 - A321836 for similar sequences.

%K sign,mult

%O 1,2

%A _N. J. A. Sloane_, Nov 23 2018