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A099622
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1)*(5/4)^k.
2
0, 1, 8, 53, 316, 1785, 9744, 51997, 273092, 1417889, 7299160, 37334661, 190028748, 963565513, 4871514656, 24572321645, 123720601684, 622038982257, 3123938806632, 15674669614549, 78593250398300, 393845861293721
OFFSET
0,3
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).
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)
MATHEMATICA
LinearRecurrence[{8, -11, -20}, {0, 1, 8}, 30] (* Harvey P. Dale, Nov 05 2017 *)
PROG
(Magma) [(5^(n+2) -6*4^(n+1) -(-1)^n)/30: n in [0..40]]; // G. C. Greubel, Jul 22 2022
(SageMath) [(5^(n+2) -6*4^(n+1) -(-1)^n)/30 for n in (0..40)] # G. C. Greubel, Jul 22 2022
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Oct 25 2004
STATUS
approved