A113252 Corresponds to m = 6 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, 92, 784, -3856, 33856, 96704, 73984, -418048, 59474944, -101917696, 443355136, 6249181184, 37406654464, -217868812288, 2345945595904, 4101714673664, 699056521216, 52661959000064, 3420344569298944, -8264891921072128, 41548867031793664

Offset

0,2

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 (-4,0,144,1296).

Formula

G.f.: (-1+108*x^2+1296*x^3)/((6*x+1)*(1-6*x)*(36*x^2+4*x+1)).

a(n) = -4*a(n-1) + 144*a(n-3) + 1296*a(n-4) for n>3. - Colin Barker, May 20 2019

Prog

(PARI) Vec(-(1 - 108*x^2 - 1296*x^3) / ((1 - 6*x)*(1 + 6*x)*(1 + 4*x + 36*x^2)) + O(x^25)) \\ Colin Barker, May 20 2019

Crossrefs

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113253, A113254, A113255, A113256.

Sequence in context: A129434 A248041 A209984 * A003737 A362511 A265238

Adjacent sequences: A113249 A113250 A113251 * A113253 A113254 A113255

Keyword

easy,sign

Author

Creighton Dement, Nov 18 2005

Status

approved