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Define the sequence $a_1,a_2,\cdots$ by $2^n = \sum_{d | n} a_d.$ Show $a_n$ is divisible by n for every $n\ge 1$.

A solution I found online defined the following sequence $b_n$ as follows. Let $b_n$ be the number of sequences of length n made up of $0$s and $1$s that can't be divided into more than one identical block (i.e. it can't be written as the concatenation of more than one identical block).

Why does $b_n$ satisfy the same recurrence relation as $a_n$? Computing the first few values of $a_n$ gives $a_1=2,a_2=2,a_3=6$. Also, $b_1 = 2, b_2 = 2, b_3 = 6.$

Also the solution claimed that $b_n$ is divisible by $n$ because the non-repeating sequences of length n can be divided into groups of n, where the sequences are obtained from each other by cyclic shifts.

To justify this, I suppose the standard approach is to define an equivalence relation on the non-repeating sequences of length n by stating that one sequence is equivalent to another iff they are cyclic shifts of each other. Clearly this is an equivalence relation, and it is known that equivalence relations are disjoint. Thus the only issue is proving that each equivalence class has n elements. To justify this, one approach might be to show that if two distinct cyclic shifts equal each other, then the sequence must have an identical block. So let $B_1\cdots B_n$ and $B_1'\cdots B_n'$ be cyclic shifts of each other so that for all i, $B_i' = B_{i+k}$ for some $k < n$, where indices are taken modulo $n$. We have $B_i = B_{i+k}$ for all k, but I'm not sure how to show there must be at least two identical blocks.

user3379
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    For your further reading, see the duplicates, which also contain some powerful generalisations of the results here. – Parcly Taxel Sep 24 '22 at 15:58

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Classify cyclic binary sequences of length $n$ according to their shortest period. If this period is $d$, then $d|n$. Thus the set of cyclic binary sequences is partitioned by this classification, yielding $2^n=\sum_{d|n} b_d$.

Yuval Peres
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  • So to formalize this better and check my understanding, I'll describe what I understand as follows. Every binary sequence of length n has a shortest period dividing n. Indeed, one can use indices modulo n so that n is a period of every such sequence. Then to justify that the shortest period must divide n, write $n = qd + r$ for some $0\leq r < d.$ $d$ is a period, so if $a_i$ is an element of a given binary sequence, we have $a_{i+qd} = a_i.$ Also $n$ is a period, so $a_{i+n}=a_i.$ Then $a_{i+r} = a_{i+n}$ as $d$ is a period, so $a_{i+r} = a_i.$ – user3379 Sep 24 '22 at 15:21
  • Hence r is a period, and by the minimality of $d, r =0$. Now to justify the second part of the equation, we have $B_{i+k} = B_i$ for all $i$, meaning that $k$ is a period of the sequence. From above, $k | n.$ But then since $k<n$, the sequence $B_1\cdots B_n$ can be split into $n/k > 1$ groups of an identical block of length k, yielding the desired result. – user3379 Sep 24 '22 at 15:23
  • There is further explanation in the duplicates. – Yuval Peres Sep 25 '22 at 00:30