A108475 Expansion of (1-3*x) / (1-5*x-5*x^2+x^3).

1, 2, 15, 84, 493, 2870, 16731, 97512, 568345, 3312554, 19306983, 112529340, 655869061, 3822685022, 22280241075, 129858761424, 756872327473, 4411375203410, 25711378892991, 149856898154532, 873430010034205, 5090723162050694, 29670908962269963

Offset: 0

Author

Paul Barry, Jun 04 2005

Keywords

• easy
• nonn

Comments

Row sums of A108477. In general, Sum_{k=0..n} Sum_{j=0..n} binomial(2*(n-k), j)*binomial(2*k, j)*r^j has expansion (1 - (r+1)*x)/(1 - (r+3)*x - (r-1)*(r+3)*x^2 + (r-1)^3*x^3).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Programs

Mathematica

CoefficientList[Series[(1-3x)/(1-5x-5x^2+x^3),{x,0,30}],x] (* or *) LinearRecurrence[{5,5,-1},{1,2,15},30] (* Harvey P. Dale, Dec 30 2019 *)

PARI

Vec((1-3*x)/((1+x)*(1-6*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 04 2016

Formula

G.f.: (1-3*x)/((1+x)*(1-6*x+x^2)).

a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3).

a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(2*(n-k), j)*binomial(2*k, j)*2^j.

Conjecture: a(n) = A000129(n+1)*A001333(n). - R. J. Mathar, Jul 08 2009

a(n) + a(n+1) = A001541(n+1). - R. J. Mathar, Jul 13 2009

a(n) = (4*(-1)^n - (3-2*sqrt(2))^n*(-2+sqrt(2)) + (2+sqrt(2))*(3+2*sqrt(2))^n)/8. - Colin Barker, Nov 04 2016

a(n) = (-1)^n * Re(sqrt(1+i) * cos((n + 1/2) * arccos(i)) * sin(n * arccos(i)) + 1), where i = sqrt(-1). - Daniel Suteu, Jun 23 2018