A249638 Number of strings of length n over a 5-letter alphabet that begin with a nontrivial palindrome.

0, 0, 5, 45, 245, 1305, 6605, 33405, 167405, 838845, 4196045, 20989245, 104955245, 524820945, 2624149445, 13120970445, 65605075445, 328026491505, 1640133571805, 8200673428605, 41003372712605, 205016891401905, 1025084484848405, 5125422563427405

Offset: 0

Author

Peter Kagey, Nov 02 2014

Keywords

• easy
• nonn
• walk

Comments

A nontrivial palindrome is a palindrome of length 2 or greater. (E.g., "1" is a trivial palindrome, but "11" and "121" are nontrivial palindromes.)

For example, 0042 is a string in a five letter alphabet of length 4 that begins with a nontrivial palindrome (00).

5 divides a(n) for all n.

Number of walks of n steps that begin with a palindromic sequence on the complete graph K_5 with loops. (E.g., 0, 1, 1, 0, 4, 1, 2 is a valid walk with 7 steps and begins with the palindromic sequence '0110'.)

Limit_{n->oo} a(n)/5^n ~ 0.429951613027098 is the probability that a random, infinite string in a five letter alphabet begins with a nontrivial palindrome.

Examples

			For n=3 the a(3) = 45 valid strings are: 000, 001, 002, 003, 004, 010, 020, 030, 040, 101, 110, 111, 112, 113, 114, 121, 131, 141, 202, 212, 220, 221, 222, 223, 224, 232, 242, 303, 313, 323, 330, 331, 332, 333, 334, 343, 404, 414, 424, 434, 440, 441, 442, 443, 444.
		

Links

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

Crossrefs

Analogous sequences for k-letter alphabets: A248122 (k=3), A249629 (k=4), A249639 (k=6), A249640 (k=7), A249641 (k=8), A249642 (k=9), A249643 (k=10).

Programs

Haskell

import Data.Ratio
a 0 = 0; a 1 = 0;
a n = 5 * a(n - 1) + 5^ceiling(n % 2) - a(ceiling(n % 2)) -- Peter Kagey, Aug 13 2015

Mathematica

a249638[n_] := Block[{f},
  f[0] = f[1] = 0;
  f[x_] := 5*f[x - 1] + 5^Ceiling[x/2] - f[Ceiling[x/2]];
Table[f[i], {i, 0, n}]]; a249638[23] (* Michael De Vlieger, Dec 27 2014 *)

Ruby

seq = [0, 0]; (2..N).each{ |i| seq << 5 * seq[i-1] + 5**((i+1)/2) - seq[(i+1)/2] }

Formula

a(0) = 0; a(1) = 0; a(n+1) = 5*a(n) + 5^ceiling((n+1)/2) - a(ceiling((n+1)/2)).