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%I #37 Jul 02 2023 14:07:49
%S 3,3,9,36,108,351,1053,3240,9720,29403,88209,265356,796068,2390391,
%T 7171173,21520080,64560240,193700403,581101209,1743362676,5230088028,
%U 15690441231,47071323693,141214502520,423643507560
%N Number of reversible strings with n beads of 3 colors. If more than 1 bead, not palindromic.
%H Andrew Howroyd, <a href="/A032086/b032086.txt">Table of n, a(n) for n = 1..200</a>
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, 3, -9).
%F "BHK" (reversible, identity, unlabeled) transform of 3, 0, 0, 0, ...
%F Conjectures from _Colin Barker_, Apr 02 2012: (Start)
%F a(n) = 3*a(n-1) + 3*a(n-2) - 9*a(n-3) for n > 4.
%F G.f.: 3*x*(1 - 2*x - 3*x^2 + 9*x^3)/((1 - 3*x)*(1 - 3*x^2)).
%F (End)
%F Conjectures from _Colin Barker_, Mar 09 2017: (Start)
%F a(n) = (2*3^n - 2*3^(n/2)) / 4 for n > 2 and even.
%F a(n) = (2*3^n - 2*3^((n+1)/2)) / 4 for n > 2 and odd.
%F (End)
%F 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
%F 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
%t Join[{3}, LinearRecurrence[{3, 3, -9}, {3, 9, 36}, 24]] (* _Jean-François Alcover_, Oct 11 2017 *)
%o (PARI) a(n) = if(n<2, [3][n], (3^n - 3^(ceil(n/2)))/2); \\ _Andrew Howroyd_, Oct 10 2017
%Y Column 3 of A293500 for n>1.
%Y Cf. A032120.
%K nonn
%O 1,1
%A _Christian G. Bower_
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