A099621 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k,k+1) * 3^(n-k-1)*(4/3)^k.

0, 1, 6, 31, 144, 637, 2730, 11467, 47508, 194953, 794574, 3222583, 13023192, 52491349, 211161138, 848231779, 3403688796, 13647040225, 54685016022, 219030629455, 876994213920, 3510591943981, 14050213040826, 56224387958011

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

Cf. A094705, A099622.

Sequence in context: A291396 A012714 A094951 * A291002 A268401 A346226

Adjacent sequences: A099618 A099619 A099620 * A099622 A099623 A099624

Keyword

easy,nonn

Author

Paul Barry, Oct 25 2004

Status

approved