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Let $n$ be a positive natural number. A sequence of $n$ positive positive integers (not necessarily distinct) is called a ”four-group” sequence if it satisfies the following requirements: for any natural $k \geq 2$, if $k$ occurs in this sequence, then $k − 1$ occurs, and the first position at which the number $k − 1$ occurs, is earlier than the last position where the number $k$ occurs. Find the number of ”four-group” sequences of length $n$.

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  • $\begingroup$ What have you tried? Have you listed them for small $n$? What is the greatest number that can occur in a sequence of length $n$? $\endgroup$ – Ross Millikan Apr 19 at 18:23
  • $\begingroup$ Do you mean integer digits or integers? The way the question is raised the answer is infinity. $\endgroup$ – Joker123 Apr 19 at 18:27
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    $\begingroup$ @Joker123 The rules imply that only integers between $1$ and $n$ can occur in the the sequence (if $n+1$ appeared, then so would $n$, and therefore also $n-1$, and so on down to $1$, which is too many numbers). Therefore, there are at most $n^n$ sequences. $\endgroup$ – Mike Earnest Apr 19 at 18:28
  • $\begingroup$ $$\sum k^{n-k} \binom nk$$ $\endgroup$ – Matt Samuel Apr 19 at 18:41
  • $\begingroup$ @MattSamuel: I don't see why it should be that. Presumably $k$ is the highest number in the sequence. If $k=2$ we count all strings of $1$s and $2$s except those where all the $2$s precede all the $1$s (and those with just one number present), so there are $2^n-n-1$ $\endgroup$ – Ross Millikan Apr 19 at 19:03
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There are $n!$ four-group sequences of length $n$.

Here is a bijection between FGS's and permutations. Given a permutation of length $n$, partition it into maximal decreasing consecutive subsequences. For example, with $\pi=[\pi_1,\pi_2,\dots,\pi_{n}]$ in one-line notation, and $n=10$, $$ [7,5,2,1, 3,10,6,9,8,4]\implies 7,5,2,1\def\div{\;\Big|\;}\div3\div10,6\div9,8,4 $$ To make the FGS corresponding to this permutation, we place a $1$ at each index in the first subsequence (at spots $7,5,2$ and $1$), we place a $2$ at each index in the second decreasing subsequence (at spot $3$), and in general place the number $k$ at each position appearing in the $k^{th}$ block. The result in this example is $$ 1, 1, 2, 4, 1, 3, 1, 4, 4, 3 $$ Note that the set of numbers appearing will always be a consecutive interval of integers containing $1$. Furthermore, the maximality of the decreasing subsequences implies that the first instance of $k-1$ is before the last instance of $k$, for each $k$.

The inverse map is as follows. Given an FGS, arrange the indices of the appearances of $1$ in decreasing order, then append the indices of the appearances of $2$ in decreasing order, and so on. I leave it to you to verify these maps are inverse to each other.

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