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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. 1
10, 11001, 1110100011, 1111011001010000111, 111110111001101011000101001000001111, 111111011110011101011100011011010011001011000010101000100100000011111 (list; graph; refs; listen; history; text; internal format)



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]


Table of n, a(n) for n=1..6.

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


a(n) = A004086(A327232(n))*10^(n-2) + A002275(n-2) for n > 1.

a(n) = A007088(A166316(n))*10^(n-1) + A002275(n-1).

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


Cf. A007088, A166315 (earliest binary de Bruijn sequences), A166316 (largest binary de Bruijn sequences), A327232 (smallest k).

Sequence in context: A267705 A086164 A138119 * A138118 A267130 A267850

Adjacent sequences:  A327230 A327231 A327232 * A327234 A327235 A327236




Jinyuan Wang, Oct 26 2019



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Last modified February 18 03:33 EST 2020. Contains 332006 sequences. (Running on oeis4.)