OFFSET
0,2
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
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).
FORMULA
G.f.: (-1+27*x^2+81*x^3)/((-3*x+1)*(3*x+1)*(9*x^2+4*x+1)).
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
EXAMPLE
a(3, 13) = 93161710957356599364/((-2+i*sqrt(5))^14*(2+i*sqrt(5))^14) = 4072324 = 2^2*1009^2.
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 *)
PROG
(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
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Creighton Dement, Oct 20 2005
STATUS
approved