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I recently realized that the number of unimodal cyclic permutations of ${1,2,\dots,n}$ matched the number of primitive binary necklaces where complements are considered equivalent, given by the formula $$\frac{1}{2n}\sum_{d \mid n, d\text{ odd}}\mu(d)2^{n/d}$$ as described in A000048. I do not have a proof of this, but it seems to hold for all $1 \le n \le 20$. My question is twofold:

  1. Why does this formula apply for primitive binary necklaces where complements are equivalent? In particular, I know that one can derive that the number of primitive binary strings of length $n$ is $\sum_{d \mid n}\mu(d)2^{n/d}$ using Möbius inversion. Dividing through by $n$ accounts for rotational symmetry. Halving the count accounts for complement symmetry, but there are some cases where a rotation might map a necklace to its complement. But then how does d being odd resolve this?
  2. Does there exists a bijection between unimodal cyclic permutations and primitive binary necklaces where complements are equivalent? If not, then how can one prove my original assertion that the number of unimodal cyclic permutations does indeed follow the above formula? A simple proof requiring little higher math, if possible, would be appreciated.

As an example of what I mean by "unimodal cyclic permutations", the three such permutations for $\{1,2,3,4,5\}$ are $(2,4,5,3,1)$, $(2,3,4,5,1)$, and $(3,5,4,2,1)$, which form the cycles $(1\:2\:4\:3\:5)$, $(1\:2\:3\:4\:5)$, and $(1\:3\:4\:2\:5)$, respectively. The $3$ corresponding primitive binary necklaces, expressed in Lyndon word form, would be $00001$, $00011$, and $00101$.

japple
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  • What does "unimodal" mean for a cyclic permutation? Same as in https://arxiv.org/abs/math/0102051 ? – darij grinberg Dec 27 '24 at 16:54
  • Yes, as in increasing up to a point and then decreasing. The paper you linked does provide the above formula for the number of unimodal cyclic permutations, but unfortunately the paper they cited for this formula is not publicly available. – japple Dec 27 '24 at 17:04
  • The paper "Counting permutations with given cycle structure and descent set" by Gessel and Reutenauer may be relevant. – Jair Taylor Dec 27 '24 at 20:25
  • From what I can understand, Theorem 2.1 in the paper you mentioned does prove my assertion (it is also cited as a part of the proof of Theorem 1.1 in the paper darij linked, which likewise seems to prove the assertion). However, the proofs and usage of these two theorems are quite out of my depth, while proving the same formula applies for primitive binary necklaces where complements are equivalent seems much more manageable (except for the part about odd d). I was hoping that there would be a relatively simple bijection, or there would be some simpler proof of the assertion. – japple Dec 27 '24 at 22:39
  • How do you derive that the number of primitive binary strings of length n is $\sum_{d|n} \mu(d) 2^{n/d}$ using Möbius inversion? It seems you need to argue $2^n = \sum_{d|n} f(d)$ where $f$ is this desired number but I don't see why this is true. – DuduBob Dec 27 '24 at 22:59
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    There are $2^n$ binary strings of length $n$, and by definition every binary string is made up of $\frac{n}{d}$ repetitions of some length $d$ primitive binary string, so we have $2^n=\sum_{d \mid n}f(d)$ (where $f(d)$ is the number of primitive binary strings). Then Möbius inversion finishes. – japple Dec 27 '24 at 23:49
  • Some progress on part 1 of the question: it suffices to prove that the number of self-complementary primitive binary necklaces of length $n$ is $$\frac{1}{n}\sum_{\substack{d \mid \frac{n}{2} \ d \text{ odd}}}\mu(d)2^{n/(2d)}.$$ This would also be the number of primitive binary necklaces of length $\frac{n}{2}$ where complements are considered equivalent. – japple Dec 30 '24 at 01:48

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