A032086 Number of reversible strings with n beads of 3 colors. If more than 1 bead, not palindromic.
3, 3, 9, 36, 108, 351, 1053, 3240, 9720, 29403, 88209, 265356, 796068, 2390391, 7171173, 21520080, 64560240, 193700403, 581101209, 1743362676, 5230088028, 15690441231, 47071323693, 141214502520, 423643507560
Offset: 1
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..200
- C. G. Bower, Transforms (2)
- Index entries for linear recurrences with constant coefficients, signature (3, 3, -9).
Programs
-
Mathematica
Join[{3}, LinearRecurrence[{3, 3, -9}, {3, 9, 36}, 24]] (* Jean-François Alcover, Oct 11 2017 *) -
PARI
a(n) = if(n<2, [3][n], (3^n - 3^(ceil(n/2)))/2); \\ Andrew Howroyd, Oct 10 2017
Formula
"BHK" (reversible, identity, unlabeled) transform of 3, 0, 0, 0, ...
Conjectures from Colin Barker, Apr 02 2012: (Start)
a(n) = 3*a(n-1) + 3*a(n-2) - 9*a(n-3) for n > 4.
G.f.: 3*x*(1 - 2*x - 3*x^2 + 9*x^3)/((1 - 3*x)*(1 - 3*x^2)).
(End)
Conjectures from Colin Barker, Mar 09 2017: (Start)
a(n) = (2*3^n - 2*3^(n/2)) / 4 for n > 2 and even.
a(n) = (2*3^n - 2*3^((n+1)/2)) / 4 for n > 2 and odd.
(End)
The above conjectures are true: The second set follows from the definition and the first set can be derived from that. - Andrew Howroyd, Oct 10 2017
a(n) = (3^n - 3^(ceiling(n/2))) / 2 = (A000244(n) - A056449(n)) / 2 for n>1. - Robert A. Russell and Danny Rorabaugh, Jun 22 2018