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It is widely known considering the set $\{1,2,\cdots,\ n\}$ that $ \lim_{n\to\infty}\frac{n!}{!n}=e$ where $!n$ are the derangements.

Less known is the fact that defining $a_n$ as the number of permutations such that no consecutive digits have adjacent positions (Hertzsprung's problem, see A002464), $ \lim_{n\to\infty}\frac{n!}{a_n}=e^2$ (see Number of $9-$digit numbers, all digits are different and nonzero and having no consecutive digits in consecutive positions).

Is it possible to find examples of subsets of the permutations of $\{1,2,\cdots,\ n\}$, such that defining their number $b_n$, holds $ \lim_{n\to\infty}\frac{n!}{b_n}=e^k$ with the integer $k>2$?

Update:

One possible case $k=3$ is discussed here (simply considering the derangements among the permutations of the Hertzsprung's problem) : If some series of n terms is deranged, what is the probability that no term stands next to a term it was next to originally?

https://oeis.org/A288208

user967210
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  • It would reasonable to consider permutations such that adjacent digits are at distance at least 3 (on a circular representation). But I did not try to prove it. – Eric Brier Feb 21 '25 at 18:58
  • Very roughly: the number of fixed points of a permutation is Poisson(1)-distributed in the limit. The number of consecutive digits in consecutive positions is probably Poisson(2)-distributed in the limit (any consecutive pair of positions has probability 2/n of being consecutive and there are n such pairs). The example of @EricBrier probably works with $k = 4$. – Michael Lugo Feb 21 '25 at 20:36
  • @EricBrier do you know how to compute them? – user967210 Feb 22 '25 at 14:06
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    I had this question about generalizing the derangement problem to blocks of size $k$: https://math.stackexchange.com/q/4913547/111594 (I used the term multi-permutation, but we can consider it to be regular permutations and "forbidden blocks" of size $k$). – ploosu2 Feb 22 '25 at 21:21
  • @ploosu2 very interesting! So what could be a good example with $ k=3$? – user967210 Feb 22 '25 at 22:09
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    @user967210 That would be the permutations $\sigma \in S_{3n}$ with $\sigma(1), \sigma(2), \sigma(3) \notin {1,2,3}, \sigma(4), \sigma(5), \sigma(6) \notin {4,5,6}$ and so on. I guess another way of forming this is that $\sigma(j) \not\equiv j \pmod 3 $. (That is we "spread the blocks".) And if $n$ is not a multiple of $3$, I think the rest couple of elements don't matter that much in the limit. – ploosu2 Feb 23 '25 at 07:50

1 Answers1

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Here is a way to do it which doesn't look anything like a generalization of Hertzsprung's problem, but it is a generalization of the derangement count. A corollary of the exponential formula in permutation form is that if $c_1 < c_2 < \dots < c_i$ is a sequence of distinct positive integers then the asymptotic density of the number of permutations $a_n$ which do not have cycles of any length $c_i$ is

$$\lim_{n \to \infty} \frac{a_n}{n!} = \exp \left( - \sum_{j=1}^i \frac{1}{c_j} \right).$$

The sequence $c_j$ can even be infinite. So for any particular positive integer $k$ the problem reduces to finding a sequence $c_i$ satisfying

$$k = \sum_{j=1}^i \frac{1}{c_j}.$$

This is the Egyptian fraction problem and is always solvable (although sources seem to only prove this for positive rationals between $0$ and $1$ it is actually true for all positive rationals); for example

$$2 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$$ $$3 = \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \right) + \frac{1}{4} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \right) + \left( \frac{1}{5} + \frac{1}{20} + \frac{1}{13} + \frac{1}{156} + \frac{1}{7} + \frac{1}{42} \right)$$

where that mess on the right was generated by starting with $\frac{1}{2}$ and then repeatedly applying the transformation

$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}.$$

I also gave this problem to GPT-o3-mini who used a greedy algorithm to find

$$3 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{15} + \frac{1}{230} + \frac{1}{57960}.$$

To keep going from here you can add a new term $1$ and repeatedly apply the above identity whenever you get a denominator you've gotten already until your denominators are all distinct and new. Incidentally, this argument implies that the harmonic series diverges.

Admittedly this is a bit artificial although I think the $k = 2$ case is nice; we are counting permutations $g$ with no fixed points, $2$-cycles, $3$-cycles, or $6$-cycles, or equivalently permutations such that $g^6$ has no fixed points. If we use infinite sums we could instead go with

$$2 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots $$

and then we'd be counting permutations with no cycles of length a power of $2$, or equivalently such that no power $g^{2^k}$ has a fixed point.

Qiaochu Yuan
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    You could replace 6 with any perfect number. Similarly the probability that $g^{120}$ has no fixed points will comes out to be $e^{-3}$ since 120 is a "3-perfect number", i.e. the sum of its divisors is 3 times itself. But this seems more artificial than the sixth power. See https://en.wikipedia.org/wiki/Multiply_perfect_number – Michael Lugo Feb 21 '25 at 20:41
  • Wow! Great answer, the cases with $k = 2$ are really nice. I admit I was looking maybe for $k=3$ for something with a more combinatorial flavour. I add this sequence for $k=2$ https://oeis.org/A003471 – user967210 Feb 21 '25 at 20:44