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).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-5,-12).
FORMULA
G.f.: x^2/((1-3*x)*(1-3*x-4*x^2)).
a(n) = 6*a(n-1) - 5*a(n-2) - 12*a(n-3).
From G. C. Greubel, Jun 06 2019: (Start)
a(n) = (4^(n+2) - 5*3^(n+1) - (-1)^n)/20.
E.g.f.: (-exp(-x) - 15*exp(3*x) + 16*exp(4*x))/20. (End)
MATHEMATICA
Table[Sum[Binomial[n-k, k+1]3^(n-k-1) (4/3)^k, {k, 0, Floor[n/2]}], {n, 0, 25}] (* or *) LinearRecurrence[{6, -5, -12}, {0, 1, 6}, 30] (* Harvey P. Dale, Dec 13 2012 *)
Table[(4^(n+2)-5*3^(n+1)-(-1)^n)/20, {n, 0, 30}] (* G. C. Greubel, Jun 06 2019 *)
PROG
(PARI) vector(30, n, n--; (4^(n+2)-5*3^(n+1)-(-1)^n)/20) \\ G. C. Greubel, Jun 06 2019
(Magma) [(4^(n+2)-5*3^(n+1)-(-1)^n)/20: n in [0..30]]; // G. C. Greubel, Jun 06 2019
(Sage) [(4^(n+2)-5*3^(n+1)-(-1)^n)/20 for n in (0..30)] # G. C. Greubel, Jun 06 2019
(GAP) List([0..30], n-> (4^(n+2)-5*3^(n+1)-(-1)^n)/20) # G. C. Greubel, Jun 06 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Oct 25 2004
STATUS
approved