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Fourth convolution of A000129(n+1) (generalized (2,1)-Fibonacci, called Pell numbers), n>=0, with itself.
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%I #24 Oct 03 2022 08:45:27

%S 1,10,65,340,1555,6482,25235,93200,330070,1129580,3756950,12197320,

%T 38787770,121148300,372476410,1129367632,3382133695,10016694470,

%U 29370557375,85341915260,245939376949,703423066190

%N Fourth convolution of A000129(n+1) (generalized (2,1)-Fibonacci, called Pell numbers), n>=0, with itself.

%H G. C. Greubel, <a href="/A073381/b073381.txt">Table of n, a(n) for n = 0..1000</a>

%H Milan Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Janjic/janjic33.html">Hessenberg Matrices and Integer Sequences </a>, J. Int. Seq. 13 (2010) # 10.7.8.

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-35,40,30,-68,-30,40,35,10,1).

%F a(n) = Sum_{k=0..n} b(k)*c(n-k), with b(k) = A000129(k+1) and c(k) = A073380(k).

%F a(n) = Sum_{k=0..floor(n/2)} 2^(n-2*k) * binomial(n-k+4, 4) * binomial(n-k, k).

%F a(n) = ((2457 +2128*n +572*n^2 +48*n^3)*(n+1)*U(n+1) + 5*(123 +142*n +44*n^2 +4*n^3) *(n+2)*U(n))/(3*2^11), with U(n) = A000129(n+1), n>=0.

%F G.f.: 1/(1-(2+x)*x)^5.

%F a(n) = F''''(n+5, 2)/4!, that is, 1/4! times the 4th derivative of the (n+5)th Fibonacci polynomial evaluated at x=2. - _T. D. Noe_, Jan 19 2006

%t CoefficientList[Series[1/(1-2*x-x^2)^5, {x,0,40}], x] (* _G. C. Greubel_, Oct 02 2022 *)

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/(1-2*x-x^2)^5 )); // _G. C. Greubel_, Oct 02 2022

%o (SageMath)

%o def A073381_list(prec):

%o P.<x> = PowerSeriesRing(ZZ, prec)

%o return P( 1/(1-2*x-x^2)^5 ).list()

%o A073381_list(40) # _G. C. Greubel_, Oct 02 2022

%Y Fifth (m=4) column of triangle A054456, A073380.

%Y Cf. A000129.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Aug 02 2002