A327233 - OEIS

%I #29 Nov 21 2019 13:13:50

%S 10,11001,1110100011,1111011001010000111,

%T 111110111001101011000101001000001111,

%U 111111011110011101011100011011010011001011000010101000100100000011111

%N 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.

%C 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]

%H Jianing Song, <a href="/A327232/a327232.txt">A general proof that the first and last n-1 digits of a(n) are identical</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/deBruijnSequence.html">de Bruijn Sequence</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/De_Bruijn_sequence">de Bruijn Sequence</a>

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

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

%F 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

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

%K nonn

%O 1,1

%A _Jinyuan Wang_, Oct 26 2019