A327319 a(n) = binomial(n, 2) + 6*binomial(n, 4).
0, 0, 1, 3, 12, 40, 105, 231, 448, 792, 1305, 2035, 3036, 4368, 6097, 8295, 11040, 14416, 18513, 23427, 29260, 36120, 44121, 53383, 64032, 76200, 90025, 105651, 123228, 142912, 164865, 189255, 216256, 246048, 278817, 314755, 354060, 396936
Offset: 0
Examples
a(5) = binomial(5, 2) + 6*binomial(5, 4) = 10 + 6*5 = 40.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
-
Mathematica
Table[Binomial[n, 2] + 6Binomial[n, 4], {n, 0, 39}] (* Alonso del Arte, Sep 18 2019 *) LinearRecurrence[{5,-10,10,-5,1},{0,0,1,3,12},40] (* Harvey P. Dale, Dec 10 2022 *) -
PARI
a(n) = {binomial(n, 2) + 6 * binomial(n, 4)} \\ Andrew Howroyd, Sep 20 2019 -
PARI
concat([0,0], Vec(x^2*(1 - 2*x + 7*x^2) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Sep 25 2019
Formula
From Colin Barker, Sep 21 2019: (Start)
G.f.: x^2*(1 - 2*x + 7*x^2) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
a(n) = (n*(-8 + 13*n - 6*n^2 + n^3)) / 4. (End)
E.g.f.: (1/4)*exp(x)*x^2*(2 + x^2). - Stefano Spezia, Sep 21 2019
Comments