A110311 Expansion of 1/((1+x+x^2)*(1+5*x+x^2)).

1, -6, 29, -138, 660, -3162, 15151, -72594, 347819, -1666500, 7984680, -38256900, 183299821, -878242206, 4207911209, -20161313838, 96598657980, -462831976062, 2217561222331, -10624974135594, 50907309455639, -243911573142600, 1168650556257360, -5599341208144200

Offset: 0

Author

Creighton Dement, Jul 19 2005

Keywords

• easy
• sign

Comments

In reference to the program code, A004254(n+1) = 1ibaseiseq[A*B](n).

Superseeker finds: a(n) + a(n+1) + a(n+2) = (-1)^n*A004254(n+3).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-6,-7,-6,-1).

Crossrefs

Cf. A004253, A004254, A110307, A110308, A110309, A110310.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1+x+x^2)*(1+5*x+x^2)) )); // G. C. Greubel, Jan 02 2023

Maple

seriestolist(series(1/((x^2+5*x+1)*(x^2+x+1)), x=0,25));

Mathematica

LinearRecurrence[{-6,-7,-6,-1}, {1,-6,29,-138}, 40] (* G. C. Greubel, Jan 02 2023 *)

PARI

Vec(1/((1+x+x^2)*(1+5*x+x^2)) + O(x^25)) \\ Colin Barker, May 14 2019

SageMath

def U(n,x): return chebyshev_U(n,x)
def A110311(n): return (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2))
[A110311(n) for n in range(41)] # G. C. Greubel, Jan 02 2023

Formula

a(n+2) = - 5*a(n+1) - a(n) + ((-1)^n)*A109265(n+1)/2.

a(n) = -6*a(n-1) - 7*a(n-2) - 6*a(n-3) - a(n-4) for n>3. - Colin Barker, May 14 2019

a(n) = (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2)), where U(n, x) = ChebyshevU(n, x). - G. C. Greubel, Jan 02 2023