A113256 Corresponds to m = 10 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, 284, 8464, -42256, 322624, 4935104, 47997184, -485499136, 7142278144, 39980801024, 125848981504, -2501476028416, 97421005963264, 60463578988544, 16045087719424, 13889461750267904, 942837644226985984, -3160296751934734336, 18357422585040338944

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..900
• Index entries for linear recurrences with constant coefficients, signature (-4,0,400,10000).

Crossrefs

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113254, A113255.

Programs

Mathematica

LinearRecurrence[{-4, 0, 400, 10000}, {-1, 4, 284, 8464}, 25] (* Paolo Xausa, Jun 10 2024 *)

PARI

Vec(-(1 - 300*x^2 - 10000*x^3) / ((1 - 10*x)*(1 + 10*x)*(1 + 4*x + 100*x^2)) + O(x^20)) \\ Colin Barker, May 20 2019

Formula

G.f.: (-1+300*x^2+10000*x^3) / ((10*x+1)*(1-10*x)*(100*x^2+4*x+1)).

a(n) = -4*a(n-1) + 400*a(n-3) + 10000*a(n-4) for n > 3. - Colin Barker, May 20 2019