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%I #39 Nov 09 2019 10:57:29
%S 10,10011,1000101110,1000010011010111100,
%T 100000100011001010011101011011111000,
%U 100000010000110001010001110010010110011010011110101011101101111110000
%N Smallest integer k 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) = A007088(A166315(n))*10^(n-2) + 10^(2^n+n-2) for 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 = 1, (d_2)*2^(2^n-1) + (d_3)*2^(2^n-2) + ... + (d_2^n)*2^1 + (d_1)*2^0 >= A166315(n), and d_2, d_3, ..., d_(n-1) >= 0. The equalities can hold simultaneously (when written as a 2^n-digit binary number, A166315(n) begins with n 0's and ends with a 1), which gives the formula. - _Jianing Song_, Oct 28 2019
%F a(n) = A004086(floor(A327233(n)/10^(n-2)))*10^(n-2) for n > 1. - _Jinyuan Wang_, Nov 02 2019
%e For n = 2, the set of all n consecutive digits of 10011 is {10, 00, 01, 11} and the set of 0 to 2^n-1 in binary is {00, 01, 10, 11}.
%e For n = 3, the set of all n consecutive digits of 1000101110 is {100, 000, 001, 010, 101, 011, 111, 110} and the set of 0 to 2^n-1 in binary is {000, 001, 010, 011, 100, 101, 110, 111}.
%o (PARI) a(n) = {my(v=vector(2^n), w=vector(2^n)); k=2^(2^n+n-2)-1; for(i=1, 2^n, v[i]=fromdigits(binary(i-1))); while(Set(w)!=v, u=binary(k++); w[1]=fromdigits(u)\10^(2^n-1); for(i=2, 2^n, w[i]=u[i+n-1]+10*(w[i-1]%10^(n-1)))); fromdigits(u); }
%Y Cf. A007088, A166315 (earliest binary de Bruijn sequences), A166316 (largest binary de Bruijn sequences), A327233 (largest k).
%K nonn
%O 1,1
%A _Jinyuan Wang_, Oct 26 2019
%E a(5)-a(6) from _Jinyuan Wang_, Nov 02 2019
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