A113253 Corresponds to m = 7 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, 131, 1681, -8341, 68644, 369431, 923521, -10266601, 278289124, -45142549, 385690321, 28351798019, 545917055044, -2216460177409, 15348835582081, 113677067503919, 421612384372804, -3999798649362349, 75132454060794001

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 (-4,0,196,2401).

Crossrefs

Cf. A000302, A097948, A056450, A113249, A113250, A113251, A113252, A113254, A113255, A113256.

Programs

Mathematica

LinearRecurrence[{-4, 0, 196, 2401}, {-1, 4, 131, 1681}, 25] (* Paolo Xausa, Jun 10 2024 *)

PARI

Vec(-(1 - 147*x^2 - 2401*x^3) / ((1 - 7*x)*(1 + 7*x)*(1 + 4*x + 49*x^2)) + O(x^25)) \\ Colin Barker, May 20 2019

Formula

G.f.: (-1+147*x^2+2401*x^3) / ((7*x+1)*(1-7*x)*(49*x^2+4*x+1)).

a(n) = -4*a(n-1) + 196*a(n-3) + 2401*a(n-4) for n > 3. - Colin Barker, May 20 2019

a(n) = 7^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/7)*(n+1)))/4. - Eric Simon Jacob, Jul 30 2023