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A252699 Number of strings of length n over a 6-letter alphabet that do not begin with a palindrome.

%I A252699 #17 Mar 24 2017 00:47:57
%S A252699 0,6,30,150,870,5070,30270,180750,1083630,6496710,38975190,233820870,
%T A252699 1402894950,8417188950,50502952950,303016634070,1818098720790,
%U A252699 10908585828030,65451508471470,392709011853630,2356254032146590,14137523959058670,84825143520531150
%N A252699 Number of strings of length n over a 6-letter alphabet that do not begin with a palindrome.
%C A252699 6 divides a(n) for all n.
%C A252699 lim n -> infinity a(n)/6^n ~ 0.644461670963043 is the probability that a random, infinite string over a 6-letter alphabet does not begin with a palindrome.
%C A252699 This sequence gives the number of walks on K_6 with loops that do not begin with a palindromic sequence.
%H A252699 Peter Kagey, <a href="/A252699/b252699.txt">Table of n, a(n) for n = 0..1000</a>
%F A252699 a(n) = 6^n - A249639(n) for n > 0.
%e A252699 For n = 3, the first 10 of the a(3) = 150 solutions are (in lexicographic order) 011, 012, 013, 014, 015, 021, 022, 023, 024, 025.
%t A252699 a252699[n_] := Block[{f}, f[0] = f[1] = 0;
%t A252699   f[x_] := 6*f[x - 1] + 6^Ceiling[(x)/2] - f[Ceiling[(x)/2]];
%t A252699 Prepend[Rest@Table[6^i - f[i], {i, 0, n}], 0]]; a252699[22] (* _Michael De Vlieger_, Dec 26 2014 *)
%o A252699 (Ruby) seq = [1, 0]; (2..N).each { |i| seq << 6 * seq[i-1] + 6**((i+1)/2) - seq[(i+1)/2] }; seq = seq.each_with_index.collect { |a, i| 6**i - a }
%Y A252699 A249639 gives the number of strings of length n over a 6-letter alphabet that DO begin with a palindrome.
%Y A252699 Analogous sequences for k-letter alphabets: A252696 (k=3), A252697 (k=4), A252698 (k=5), A252700 (k=7), A252701 (k=8), A252702 (k=9), A252703 (k=10).
%K A252699 easy,nonn,walk
%O A252699 0,2
%A A252699 _Peter Kagey_, Dec 20 2014