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a(n) = (-1728)^n.
0

%I #24 Dec 25 2023 18:24:44

%S 1,-1728,2985984,-5159780352,8916100448256,-15407021574586368,

%T 26623333280885243904,-46005119909369701466112,

%U 79496847203390844133441536,-137370551967459378662586974208,237376313799769806328950291431424,-410186270246002225336426103593500672

%N a(n) = (-1728)^n.

%H Caroline Nunn, <a href="https://scholar.rose-hulman.edu/rhumj/vol22/iss2/3">A Proof of a Generalization of Niven's Theorem Using Algebraic Number Theory</a>, Rose-Hulman Undergraduate Mathematics Journal: Vol. 22, Iss. 2, Article 3 (2021).

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (-1728).

%F From Caroline Nunn, p. 9: (Start)

%F a(n) = (3 + sqrt(-3))^(6*n).

%F a(n) = Sum_{k=0..3*n} (-1)^k*binomial(6*n, 2*k)*3^(6*n-k). (End)

%F O.g.f.: 1/(1 + 1728*x).

%F E.g.f.: exp(-1728*x).

%F a(n) = -1728*a(n-1) for n > 0.

%F a(n) = (-12)^(3*n).

%F a(n) = (A000244(n)*A262710(n))^3.

%t LinearRecurrence[{-1728},{1},12]

%t NestList[-1728#&,1,20] (* _Harvey P. Dale_, Dec 25 2023 *)

%Y Cf. A000244, A000578, A001021, A008585, A008588, A262710.

%K sign,easy

%O 0,2

%A _Stefano Spezia_, Dec 28 2021