A056118 a(n) = (11*n+5)*(n+4)*(n+3)*(n+2)*(n+1)/120.

1, 16, 81, 266, 686, 1512, 2982, 5412, 9207, 14872, 23023, 34398, 49868, 70448, 97308, 131784, 175389, 229824, 296989, 378994, 478170, 597080, 738530, 905580, 1101555, 1330056, 1594971, 1900486, 2251096, 2651616, 3107192, 3623312

Offset

0,2

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

a(n) = (11*n+5)*binomial(n+4,4)/5.

G.f.: (1+10*x)/(1-x)^6.

a(0)=1, a(1)=16, a(2)=81, a(3)=266, a(4)=686, a(5)=1512; for n>5, a(n) = 6*a(n-1) -15*a(n-2) +20*a(n-3) -15*a(n-4) +6*a(n-5) -a(n-6). - Harvey P. Dale, Oct 18 2013

From G. C. Greubel, Jan 17 2020: (Start)

a(n) = 11*binomial(n+5,5) - 8*binomial(n+4,4).

E.g.f.: (360 +2760*x +3720*x^2 +1560*x^3 +235*x^4 +11*x^5)*exp(x)/120. (End)

Maple

seq( (11*n+5)*binomial(n+4, 4)/5, n=0..40); # G. C. Greubel, Jan 17 2020

Mathematica

Table[((11n+5)Times@@(n+Range[4]))/120, {n, 0, 40}] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {1, 16, 81, 266, 686, 1512}, 40] (* Harvey P. Dale, Oct 18 2013 *)
Table[11*Binomial[n+5, 5] -8*Binomial[n+4, 4], {n, 0, 40}] (* G. C. Greubel, Jan 17 2020 *)

Prog

(PARI) vector(41, n, (11*n-6)*binomial(n+3, 4)/5 ) \\ G. C. Greubel, Jan 17 2020
(Magma) [(11*n+5)*Binomial(n+4, 4)/5: n in [0..40]]; // G. C. Greubel, Jan 17 2020
(Sage) [(11*n+5)*binomial(n+4, 4)/5 for n in (0..40)] # G. C. Greubel, Jan 17 2020
(GAP) List([0..40], n-> (11*n+5)*Binomial(n+4, 4)/5 ); # G. C. Greubel, Jan 17 2020

Crossrefs

Cf. A055268.

Sequence in context: A113316 A113317 A187457 * A351602 A134606 A343324

Adjacent sequences: A056115 A056116 A056117 * A056119 A056120 A056121

Keyword

nonn,easy

Author

Barry E. Williams, Jul 04 2000

Status

approved