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 (-4,0,100,625).
FORMULA
G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).
a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - Colin Barker, May 20 2019
a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - Eric Simon Jacob, Jul 29 2023
MAPLE
with(gfun): seriestolist(series((-1+75*x^2+625*x^3)/((5*x+1)*(1-5*x)*(25*x^2+4*x+1)), x=0, 25));
MATHEMATICA
LinearRecurrence[{-4, 0, 100, 625}, {-1, 4, 59, 289}, 40] (* Harvey P. Dale, Jul 05 2021 *)
PROG
(PARI) Vec(-(1 - 75*x^2 - 625*x^3) / ((1 - 5*x)*(1 + 5*x)*(1 + 4*x + 25*x^2)) + O(x^30)) \\ Colin Barker, May 20 2019
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Creighton Dement, Nov 18 2005
STATUS
approved