A113251 Corresponds to m = 5 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4*a(n-4) + (2*m)^2*a(n-3) - 4*a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6*(m-1) + 3*(m-1)^2, a(m,3) = (-8+m^2)^2.

-1, 4, 59, 289, -1381, 13924, 10079, 2209, 520439, 7628644, -23994301, 149401729, 490531859, 406344964, -1681645081, 149155846849, -249406479121, 1083427010884, 9530848465739, 30158362505569, -168169798384501, 2302905921914404, -239007146013841, 2988025311585889

Offset: 0

Author

Creighton Dement, Nov 18 2005

Keywords

• easy
• sign

Comments

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (-4,0,100,625).

Crossrefs

Cf. A000302, A097948, A056450, A113249, A113250, A113252, A113253, A113254, A113255, A113256.

Programs

Maple

with(gfun): seriestolist(series((-1+75*x^2+625*x^3)/((5*x+1)*(1-5*x)*(25*x^2+4*x+1)), x=0,25));

Mathematica

LinearRecurrence[{-4,0,100,625},{-1,4,59,289},40] (* Harvey P. Dale, Jul 05 2021 *)

PARI

Vec(-(1 - 75*x^2 - 625*x^3) / ((1 - 5*x)*(1 + 5*x)*(1 + 4*x + 25*x^2)) + O(x^30)) \\ Colin Barker, May 20 2019

Formula

G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).

a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - Colin Barker, May 20 2019

a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - Eric Simon Jacob, Jul 29 2023