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A262312
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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.
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4
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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
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0,1
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COMMENTS
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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 even-length palindrome is: lim n -> infinity a(n)/2^n ~ 0.7322131597821108... . - Peter Kagey, Jan 26 2015"
Also, the limit, as word-length approaches infinity, of the probability that a random binary word has a bifix; that is, 1-x where x is the constant from A242430. - Danny Rorabaugh, Feb 13 2016
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LINKS
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FORMULA
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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.
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EXAMPLE
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0.7322131597821108876233285964156974474449401020065154679236881114887...
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PROG
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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.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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