A113255 Corresponds to m = 9 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, 227, 5329, -26581, 206116, 2391479, 16785409, -174757993, 2826198244, 9824173259, 14210785681, -287742103741, 22876687229764, -22446053606113, 89792737665409, 5164999769137199, 122161424469552196, -606821408584323661, 4689875711360495569

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 (-4,0,324,6561).

Formula

G.f.: (-1+243*x^2+6561*x^3) / ((9*x+1)*(1-9*x)*(81*x^2+4*x+1)).

a(n) = -4*a(n-1) + 324*a(n-3) + 6561*a(n-4) for n > 3. - Colin Barker, May 20 2019

Mathematica

LinearRecurrence[{-4, 0, 324, 6561}, {-1, 4, 227, 5329}, 25] (* Paolo Xausa, Jun 10 2024 *)

Prog

(PARI) Vec(-(1 - 243*x^2 - 6561*x^3) / ((1 - 9*x)*(1 + 9*x)*(1 + 4*x + 81*x^2)) + O(x^20)) \\ Colin Barker, May 20 2019

Crossrefs

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113253, A113254, A113256.

Sequence in context: A182484 A159281 A290346 * A145767 A206352 A024057

Adjacent sequences: A113252 A113253 A113254 * A113256 A113257 A113258

Keyword

easy,sign

Author

Creighton Dement, Nov 18 2005

Status

approved