A113249 Corresponds to m = 3 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(n-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, 11, 1, 59, 484, -1009, 6241, -2761, 13924, 87251, 57121, 49139, 4072324, -7165609, 35058241, 10350959, 30492484, 559712411, 973502401, -1957852501, 30450948004, -41421000289, 174055005601, 241428053159, 9658565284, 2872244917091, 11300885699041, -25300162140061

Offset: 0

Author

Creighton Dement, Oct 20 2005

Keywords

• easy
• sign

Comments

Conjecture: a(m, 2*n+1) is a perfect square for all m, n. Disregarding signs and/or initial term, we have: m = 0 (A000302), m = 1 (A097948), m = 2 (A056450), m = 3 (a(n)), m = 4 (A113250), m = 5 (A113251), m = 6 (A113252), m = 7 (A113253), m = 8 (A113254), m = 9 (A113255), m = 10 (A113256).

In this case, a(2n+1) = b(n)^2 where b(n) = Re((2+sqrt(-5))^(n+1)) satisfies the recurrence b(n) = 4*b(n-1) - 9*b(n-2) with b(0)=2, b(1)=-1. - Robert Israel, Oct 23 2017

Examples

			a(3, 13) = 93161710957356599364/((-2+i*sqrt(5))^14*(2+i*sqrt(5))^14) = 4072324 = 2^2*1009^2.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Robert Munafo, Sequences Related to Floretions
• Index entries for linear recurrences with constant coefficients, signature (-4,0,36,81).

Crossrefs

Cf. A000302, A097948, A056450, A113250, A113251, A113252, A113253, A113254, A113255, A113256.

Cf. A176333, A190967.

Programs

Maple

f:= gfun:-rectoproc({a(n) = 81*a(n-4)+36*a(n-3)-4*a(n-1),a(0) = -1, a(1) = 4, a(2) = 11, a(3) = 1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Oct 23 2017

Mathematica

LinearRecurrence[{-4, 0, 36, 81}, {-1, 4, 11, 1}, 29] (* Jean-François Alcover, Sep 25 2017 *)

PARI

Vec(-(1 - 27*x^2 - 81*x^3) / ((1 - 3*x)*(1 + 3*x)*(1 + 4*x + 9*x^2)) + O(x^30)) \\ Colin Barker, May 19 2019

Formula

G.f.: (-1+27*x^2+81*x^3)/((-3*x+1)*(3*x+1)*(9*x^2+4*x+1)).

a(2k+1) = (2*A176333(k)-3*A190967(k))^2. - Robert Israel, Oct 23 2017

a(n) = -4*a(n-1) + 36*a(n-3) + 81*a(n-4) for n>3. - Colin Barker, May 19 2019

a(n) = 3^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/3)*(n+1)))/4. - Eric Simon Jacob, Aug 06 2023