A262312 The limit, as word-length 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 even-length palindrome.
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
Offset: 0
Examples
0.7322131597821108876233285964156974474449401020065154679236881114887...
References
- Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.17, p. 369.
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.
Programs
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Sage
N(sum([2*(1/4)^(2^j)*(-1)^j/prod([1-2*(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 j-range.
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} 1-2*(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.
Comments