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A327233
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Largest integer k < 10^(2^n+n) such that the set of all n consecutive digits of k equals the set of 0 to 2^n-1 written as n-digit binary numbers.
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1
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OFFSET
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1,1
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COMMENTS
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floor(a(n)/10^(n-1)) is the juxtaposition of a de Bruijn sequence. [This is because the first and last n-1 digits of a(n) are always identical - see my link for a general proof. - Jianing Song, Oct 29 2019]
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LINKS
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FORMULA
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Proof: by the property mentioned in the comment section, write a(n) = (d_1)*10^(2^n+n-2) + (d_2)*10^(2^n+n-3) + ... + (d_2^n)*10^(n-1) + (d_1)*10^(n-2) + (d_2)*10^(n-3) + ... + (d_(n-1))*10^0, d_i = 0 or 1, then (d_1)*2^(2^n-1) + (d_2)*2^(2^n-2) + ... + (d_2^n)*2^0 <= A166316(n), and d_1, d_2, ..., d_(n-1) <= 1. The equalities can hold simultaneously (when written as a 2^n-digit binary number, A166316(n) begins with n 1's), which gives the formula. - Jianing Song, Oct 28 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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