A099582 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1)*4^(n-k-1).

0, 0, 1, 4, 24, 112, 560, 2688, 13056, 62976, 304384, 1469440, 7096320, 34263040, 165441536, 798818304, 3857055744, 18623496192, 89922273280, 434183077888, 2096421666816, 10122418978816, 48875363631104, 235991130439680

Offset

0,4

Comments

In general a(n) = Sum{k=0..floor(n/2)} binomial(n-k, k-1) * r^(n-k-1) has g.f. x^2/((1-r*x^2)*(1-r*x-r*x^2)) and satisfies a(n) = r*a(n-1) + 2*r*a(n-2) - r^2*a(n-3) - r^2*a(n-4).

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,8,-16,-16).

Formula

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

a(n) = 4*a(n-1) + 8*a(n-2) - 16*a(n-3) - 16*a(n-4) with a(n) = (n^3-n)/6 for n<5.

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

a(n) = 2^(n-4)*(2*A000129(n) - (1 - (-1)^n)).

a(n) = (1/4)*(A057087(n-1) - 2^(n-2)*(1 - (-1)^n)).

E.g.f.: (exp(2*x)*sinh(2*sqrt(2)*x) - sqrt(2)*sinh(2*x))/(8*sqrt(2)). (End)

Mathematica

Table[Sum[Binomial[n-k, k-1]*4^(n-k-1), {k, 0, Floor[n/2]}], {n, 0, 30}] (* or *) LinearRecurrence[{4, 8, -16, -16}, {0, 0, 1, 4}, 30] (* Harvey P. Dale, Jul 19 2012 *)

Prog

(Magma) I:=[0, 0, 1, 4]; [n le 4 select I[n] else 4*(Self(n-1) +2*Self(n-2) -4*Self(n-3) -4*Self(n-4)): n in [1..41]]; // G. C. Greubel, Jul 22 2022
(SageMath) [2^(n-3)*(lucas_number1(n, 2, -1) - (n%2)) for n in (0..40)] # G. C. Greubel, Jul 22 2022

Crossrefs

Cf. A099177, A099581.

Sequence in context: A270686 A272253 A300581 * A295093 A211142 A374389

Adjacent sequences: A099579 A099580 A099581 * A099583 A099584 A099585

Keyword

easy,nonn

Author

Paul Barry, Oct 23 2004

Status

approved