A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

1, 1, 6, 7, 23, 30, 73, 103, 211, 314, 581, 895, 1560, 2455, 4135, 6590, 10890, 17480, 28590, 46070, 74946, 121016, 196326, 317342, 514123, 831465, 1346148, 2177613, 3524441, 5702054, 9227311, 14929365, 24157645, 39087010, 63245795, 102332805

Offset: 0

Author

Paul Barry, Oct 23 2004

Keywords

• easy
• nonn

Comments

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.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,5,-4,-10,6,10,-4,-5,1,1).

Crossrefs

Cf. A000045, A054450, A054451, A052952, A099571.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x^2)^4*(1-x-x^2)) )); // G. C. Greubel, Jul 25 2022

Mathematica

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 *)

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

Formula

G.f.: 1/((1-x^2)^4*(1-x-x^2)). - corrected by R. J. Mathar, Feb 20 2011

a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+3, 3)*((1+(-1)^k)/2).

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