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

Author

Aaron Kemats, Sep 17 2019

Keywords

• nonn
• easy

Comments

a(n) is the number of ternary strings of length n that have exactly two 2's, zero or two 1's, and have no restriction on the number of 0's. For example, a(6)=105 since the strings are the 90 permutations of 221100 and the 15 permutations of 220000. - Enrique Navarrete, May 19 2025

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