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

0, 6, 30, 150, 870, 5070, 30270, 180750, 1083630, 6496710, 38975190, 233820870, 1402894950, 8417188950, 50502952950, 303016634070, 1818098720790, 10908585828030, 65451508471470, 392709011853630, 2356254032146590, 14137523959058670, 84825143520531150

Offset

0,2

Comments

6 divides a(n) for all n.

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.

This sequence gives the number of walks on K_6 with loops that do not begin with a palindromic sequence.

Links

Peter Kagey, Table of n, a(n) for n = 0..1000

Formula

a(n) = 6^n - A249639(n) for n > 0.

Example

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.

Mathematica

a252699[n_] := Block[{f}, f[0] = f[1] = 0;
  f[x_] := 6*f[x - 1] + 6^Ceiling[(x)/2] - f[Ceiling[(x)/2]];
Prepend[Rest@Table[6^i - f[i], {i, 0, n}], 0]]; a252699[22] (* Michael De Vlieger, Dec 26 2014 *)

Prog

(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 }

Crossrefs

A249639 gives the number of strings of length n over a 6-letter alphabet that DO begin with a palindrome.

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).

Sequence in context: A170687 A003948 A105488 * A054117 A033132 A022023

Adjacent sequences: A252696 A252697 A252698 * A252700 A252701 A252702

Keyword

easy,nonn,walk

Author

Peter Kagey, Dec 20 2014

Status

approved