A099622 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1)*(5/4)^k.

0, 1, 8, 53, 316, 1785, 9744, 51997, 273092, 1417889, 7299160, 37334661, 190028748, 963565513, 4871514656, 24572321645, 123720601684, 622038982257, 3123938806632, 15674669614549, 78593250398300, 393845861293721

Offset: 0

Author

Paul Barry, Oct 25 2004

Keywords

• easy
• nonn

Comments

In general a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1) * u^(n-k-1) * (v/u)^(k-1) has g.f. x^2/((1-u*x) * (1-u*x-v*x^2)) and satisfies the recurrence a(n) = 2*u*a(n-1) - (u^2 - v)*a(n-2) - u*v*a(n-3).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (8,-11,-20).

Crossrefs

Cf. A094705, A099582, A099621.

Programs

Magma

[(5^(n+2) -6*4^(n+1) -(-1)^n)/30: n in [0..40]]; // G. C. Greubel, Jul 22 2022

Mathematica

LinearRecurrence[{8,-11,-20},{0,1,8},30] (* Harvey P. Dale, Nov 05 2017 *)

SageMath

[(5^(n+2) -6*4^(n+1) -(-1)^n)/30 for n in (0..40)] # G. C. Greubel, Jul 22 2022

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1)*(5/4)^k.

a(n) = 8*a(n-1) - 11*a(n-2) - 20*a(n-3).

G.f.: x^2/((1-4*x)*(1-4*x-5*x^2)) = x^2/((1+x)*(1-4*x)*(1-5*x)).

From G. C. Greubel, Jul 22 2022: (Start)

a(n) = (1/30)*(5^(n+2) - 6*4^(n+1) - (-1)^n).

E.g.f.: (1/30)*(25*exp(5*x) - 24*exp(4*x) - exp(-x)). (End)