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
MATHEMATICA
LinearRecurrence[{-4, 0, 144, 1296}, {-1, 4, 92, 784}, 25] (* Paolo Xausa, Jun 10 2024 *)
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
KEYWORD
easy,sign
AUTHOR
Creighton Dement, Nov 18 2005
STATUS
approved