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 (-4,0,196,2401).
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
MATHEMATICA
LinearRecurrence[{-4, 0, 196, 2401}, {-1, 4, 131, 1681}, 25] (* Paolo Xausa, Jun 10 2024 *)
PROG
(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
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Creighton Dement, Nov 18 2005
STATUS
approved