

A262312


The limit, as wordlength approaches infinity, of the probability that a random binary word is an instance of the Zimin pattern "aba"; also the probability that a random infinite binary word begins with an evenlength palindrome.


3



7, 3, 2, 2, 1, 3, 1, 5, 9, 7, 8, 2, 1, 1, 0, 8, 8, 7, 6, 2, 3, 3, 2, 8, 5, 9, 6, 4, 1, 5, 6, 9, 7, 4, 4, 7, 4, 4, 4, 9, 4, 0, 1, 0, 2, 0, 0, 6, 5, 1, 5, 4, 6, 7, 9, 2, 3, 6, 8, 8, 1, 1, 1, 4, 8, 8, 7, 8, 5, 0, 6, 2, 2, 1, 4, 7, 6, 7, 2, 3, 7
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OFFSET

0,1


COMMENTS

Word W over alphabet L is an instance of "aba" provided there exists a nonerasing monoid homomorphism f:{a,b}*>L* such that f(W)=aba. For example "oompaloompa" is an instance of "aba" via the homomorphism defined by f(a)=oompa, f(b)=l. For a proof of the formula or more information on Zimin words, see Rorabaugh (2015).
The second definition comes from a Comment in A094536: "The probability that a random, infinite binary string begins with an evenlength palindrome is: lim n > infinity a(n)/2^n ~ 0.7322131597821108... .  Peter Kagey, Jan 26 2015"
Also, the limit, as wordlength approaches infinity, of the probability that a random binary word has a bifix; that is, 1x where x is the constant from A242430.  Danny Rorabaugh, Feb 13 2016


LINKS

Danny Rorabaugh, Table of n, a(n) for n = 0..1000
Danny Rorabaugh, Toward the Combinatorial Limit Theory of Free Words, arXiv:1509.04372 [math.CO], 2015, University of South Carolina, ProQuest Dissertations Publishing (2015). See section 5.1.


FORMULA

The constant is Sum_{n>=0} A003000(n)*(1/4)^n.
Using the recursive definition of A003000, one can derive the series Sum_{j>=0} 2*(1)^j*(1/4)^(2^j)/(Product_{k=0..j} 12*(1/4)^(2^k)), which converges more quickly to the same limit and without having to calculate terms of A003000.
For ternary words, the constant is Sum_{n>=0} A019308(n)*(1/9)^n.
For quaternary words, the constant is Sum_{n>=0} A019309(n)*(1/16)^n.


EXAMPLE

0.7322131597821108876233285964156974474449401020065154679236881114887...


PROG

(Sage) N(sum([2*(1/4)^(2^j)*(1)^j/prod([12*(1/4)^(2^k) for k in range(j+1)]) for j in range(8)]), digits=81) #For more than 152 digits of accuracy, increase the jrange.


CROSSREFS

Cf. A003000, A094536, A123121, A242430, A262313 (abacaba).
Sequence in context: A033327 A024584 A132713 * A135296 A271353 A098907
Adjacent sequences: A262309 A262310 A262311 * A262313 A262314 A262315


KEYWORD

nonn,cons


AUTHOR

Danny Rorabaugh, Sep 17 2015


STATUS

approved



