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A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.

%I A032088 #40 Aug 06 2024 20:42:55
%S A032088 5,10,50,300,1500,7750,38750,195000,975000,4881250,24406250,122062500,
%T A032088 610312500,3051718750,15258593750,76293750000,381468750000,
%U A032088 1907347656250,9536738281250,47683710937500
%N A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.
%H A032088 Andrew Howroyd, <a href="/A032088/b032088.txt">Table of n, a(n) for n = 1..200</a>
%H A032088 C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H A032088 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5, 5, -25).
%F A032088 "BHK" (reversible, identity, unlabeled) transform of 5, 0, 0, 0, ...
%F A032088 Conjectures from _Colin Barker_, Jul 07 2012: (Start)
%F A032088 a(n) = 5*a(n-1) + 5*a(n-2) - 25*a(n-3) for n > 4.
%F A032088 G.f.: 5*x*(1 - 3*x - 5*x^2 + 25*x^3)/((1 - 5*x)*(1 - 5*x^2)).
%F A032088 (End)
%F A032088 Conjectures from _Colin Barker_, Mar 09 2017: (Start)
%F A032088 a(n) = 5^(n/2)*(5^(n/2) - 1) / 2 for n > 1 and even.
%F A032088 a(n) = -5*(5^(n/2-1/2) - 5^(n-1)) / 2 for n > 1 and odd.
%F A032088 (End)
%F A032088 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 A032088 a(n) = (5^n - 5^(ceiling(n/2))) / 2 = (A000351(n) - A056451(n)) / 2 for n>1. - _Robert A. Russell_ and _Danny Rorabaugh_, Jun 22 2018
%t A032088 Join[{5}, LinearRecurrence[{5, 5, -25}, {10, 50, 300}, 19]] (* _Jean-François Alcover_, Oct 11 2017 *)
%o A032088 (PARI) a(n) = if(n<2, [5][n], (5^n - 5^(ceil(n/2)))/2); \\ _Andrew Howroyd_, Oct 10 2017
%Y A032088 Column 5 of A293500 for n>1.
%Y A032088 Cf. A032122.
%Y A032088 Equals (A000351 - A056451) / 2 for n>1.
%K A032088 nonn,easy
%O A032088 1,1
%A A032088 _Christian G. Bower_