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%I #12 Jul 25 2022 15:42:04
%S 1,1,6,7,23,30,73,103,211,314,581,895,1560,2455,4135,6590,10890,17480,
%T 28590,46070,74946,121016,196326,317342,514123,831465,1346148,2177613,
%U 3524441,5702054,9227311,14929365,24157645,39087010,63245795,102332805
%N a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).
%C Fifth column of triangle A054450. In general Sum_{k=0..floor(n/2)} binomial(n-k+r, k), r>=0, will have g.f. 1/((1-x^2)^r*(1-x-x^2)) and for r>0, a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+r-1, r-1)*(1+(-1)^k)/2.
%H G. C. Greubel, <a href="/A099572/b099572.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,5,-4,-10,6,10,-4,-5,1,1).
%F G.f.: 1/((1-x^2)^4*(1-x-x^2)). - corrected by R. J. Mathar, Feb 20 2011
%F a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+3, 3)*((1+(-1)^k)/2).
%F a(n) = Fibonacci(n+5) + (-1)^n*(n^3 + 9*n^2 + 35*n + 33)/96 - (n^3 + 21*n^2 + 155*n + 417)/96. - _G. C. Greubel_, Jul 25 2022
%t Table[Fibonacci(n+5) +(-1)^n*(n^3+9*n^2+35*n+33)/96 -(n^3+21*n^2+155*n+417)/96, {n,0,40}] (* _G. C. Greubel_, Jul 25 2022 *)
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x^2)^4*(1-x-x^2)) )); // _G. C. Greubel_, Jul 25 2022
%o (SageMath) [fibonacci(n+5) + (-1)^n*(n^3+9*n^2+35*n+33)/96 - (n^3+21*n^2+155*n + 417)/96 for n in (0..40)] # _G. C. Greubel_, Jul 25 2022
%Y Cf. A000045, A054450, A054451, A052952, A099571.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Oct 23 2004
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