A327233 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.
10, 11001, 1110100011, 1111011001010000111, 111110111001101011000101001000001111, 111111011110011101011100011011010011001011000010101000100100000011111
Offset: 1
Keywords
Links
- Jianing Song, A general proof that the first and last n-1 digits of a(n) are identical
- Eric Weisstein's World of Mathematics, de Bruijn Sequence
- Wikipedia, de Bruijn Sequence
Crossrefs
Formula
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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