# Nested De Bruijn Sequences

A binary De Bruijn sequence of index $n$ is a circular sequence $S=a_1 a_2 \dots a_{2^n},$ with $a_i \in \{0,1\},$ and such that each of the $2^n$ binary $n$-uples occurs exactly once in $S.$

Is there an infinite binary sequence $b_1 b_2 b_3 \dots$ such that $b_1 \dots b_{2^n}$ is a binary De Bruijn Sequence of index $n$ for all $n$ ?

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WLOG the sequence starts 01. Now the first four terms have to be 0110. But this can never be part of an index 3 De Bruijn sequence, because an index 3 De Bruijn sequence has to have precisely four 1s and three of them had better be consecutive, but this can't happen with a 0110. So, sadly, the answer is "no". –  Kevin Buzzard Jun 11 '11 at 18:20
Are there at least interesting subsequences $n_1,n_2,\ldots$ of the natural numbers so that $b_1b2_\ldots b_{2^{n_i}}$ is a binary De Bruijn sequence for each $n_i$? For example, can this work if $n_i=2^i$? –  Simon Jul 9 '11 at 13:08