A135712 a(n) = (4*n^3 + 11*n^2 + 9*n + 2)/2.

1, 13, 48, 118, 235, 411, 658, 988, 1413, 1945, 2596, 3378, 4303, 5383, 6630, 8056, 9673, 11493, 13528, 15790, 18291, 21043, 24058, 27348, 30925, 34801, 38988, 43498, 48343, 53535, 59086, 65008, 71313, 78013, 85120, 92646, 100603, 109003, 117858, 127180

Offset: 0

Author

N. J. A. Sloane, Mar 05 2008

Keywords

• nonn
• easy

Comments

Binomial transform yields 1,12,23,12,0,0,0,0,0,0,.. - R. J. Mathar, Apr 21 2008

References

• J. H. Conway and R. K. Guy, The Book of Numbers, p. 83.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• M. E. Larsen, The eternal triangle - a history of a counting problem, College Math. J., 20 (1989), 370-392.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Bisection of A002717 (odd part).

Partial sums of A033570. - Bruno Berselli, Nov 28 2013

Programs

Mathematica

Table[(4*n^3 + 11*n^2 + 9*n + 2)/2,{n,0,25}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,13,48,118}, 25] (* G. C. Greubel, Oct 29 2016 *)

Formula

G.f.: (1 + 9*x + 2*x^2) / (1-x)^4. - R. J. Mathar, Apr 21 2008

From G. C. Greubel, Oct 29 2016: (Start)

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

E.g.f.: (1/2)*(2 + 24*x + 23*x^2 + 4*x^3)*exp(x). (End)

a(n) = ((2*n+1)*(2*n+3)*(4*n+3) - 1)/8 = (n+1)*(4*n^2 + 7*n + 2)/2, for n >= 0. See the Conway and Guy reference. - Wolfdieter Lang, Apr 16 2020