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%I #9 Feb 18 2023 21:16:19
%S 0,1,6,108,1080,14256,163296,2006208,23794560,287214336,3436494336,
%T 41298398208,495217981440,5944792559616,71324450021376,
%U 855971764420608,10271190988062720,123257112966660096,1479068428940476416
%N a(n) = 6^(n-1)*J(n), where J(n) = A001045(n).
%C In general k^(n-1)*J(n), where J(n) = A001045(n), is given by ((2*k)^n - (-k)^n)/(3*k) with g.f. x/((1+k*x)*(1-2*k*x)).
%H G. C. Greubel, <a href="/A099138/b099138.txt">Table of n, a(n) for n = 0..925</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,72).
%F G.f.: x/((1+6*x)*(1-12*x)).
%F a(n) = 6^(n-1)*Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k) * 2^k.
%F a(n) = (12^n - (-6)^n)/18.
%F a(n) = 6^(n-1)*A001045(n).
%F E.g.f.: (1/18)*(exp(12*x) - exp(-6*x)). - _G. C. Greubel_, Feb 18 2023
%t LinearRecurrence[{6,72}, {0,1}, 40] (* _G. C. Greubel_, Feb 18 2023 *)
%o (Magma) [(12^n - (-6)^n)/18: n in [0..40]]; // _G. C. Greubel_, Feb 18 2023
%o (SageMath) [(12^n - (-6)^n)/18 for n in range(41)] # _G. C. Greubel_, Feb 18 2023
%Y Cf. A001045, A003683, A080424, A091903, A091904.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Sep 29 2004
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