A090386 Fifth diagonal (m=4) of triangle A084938; a(n) = A084938(n+4,n) = (n^4 + 18*n^3 + 131*n^2 + 426*n)/24.

0, 24, 64, 126, 217, 345, 519, 749, 1046, 1422, 1890, 2464, 3159, 3991, 4977, 6135, 7484, 9044, 10836, 12882, 15205, 17829, 20779, 24081, 27762, 31850, 36374, 41364, 46851, 52867, 59445, 66619, 74424, 82896, 92072, 101990, 112689, 124209, 136591, 149877, 164110

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = A084938(n+4,n) = Sum_{k=0..4} A090238(4,k)*binomial(n,k).

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=24, a(2)=64, a(3)=126, a(4)=217. - Harvey P. Dale, Feb 23 2014

From Elmo R. Oliveira, May 24 2026: (Start)

G.f.: x*(24 - 56*x + 46*x^2 - 13*x^3)/(1 - x)^5.

E.g.f.: x*(12 + x)*(48 + 12*x + x^2)*exp(x)/24. (End)

Mathematica

Table[(n^4+18n^3+131n^2+426n)/24, {n, 0, 40}] (* Harvey P. Dale, Feb 23 2014 *)
(* Alternative: *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 24, 64, 126, 217}, 40] (* Harvey P. Dale, Feb 23 2014 *)

Prog

(Magma) [(n^4+18*n^3+131*n^2+426*n)/24: n in [0..40]]; // Vincenzo Librandi, Feb 24 2014

Crossrefs

Cf. A084938, A090238.

Sequence in context: A045558 A352343 A022761 * A239596 A306132 A118609

Adjacent sequences: A090383 A090384 A090385 * A090387 A090388 A090389

Keyword

easy,nonn,changed

Author

Philippe Deléham, Jan 30 2004

Extensions

Corrected by T. D. Noe, Nov 08 2006

Status

approved