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The Mompox Sequence, $a(n):=1, 2, 6, 24, 120, 20, \ldots$ (OEIS A008336), is the sequence of positive integers whose first term is $1$, and in which the $n$-th term (after the first one) equals the previous term divided by $n$ if the result is an integer, or times $n$ if not.

Is the Mompox sequence an S-sequence (S-sequences are described here), that is, are all its terms distinct?

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    $\begingroup$ @SamHopkins Nicolás Martínez, a school teacher of Santa Cruz de Mompox, Colombia, and who seemed to have been (as far as I know) the first to have, around 1986, studied it, and pose questions about it (i.e. Does it have infinitely many odd terms? ) wished it be called thus. $\endgroup$
    – Bernardo Recamán Santos
    Commented Jan 18 at 23:47
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    $\begingroup$ This sequence is indeed injective. Obviously , $a_n\ne a_{n+1}$. Note that $\nu_p(a_n)\equiv\nu_p(n!)\pmod2$ for every natural $n$ and prime $p$. In particular, if $a_n=a_{n+k}$, then $\prod_{i=1}^k(n+i)$ is a perfect square, which is not the case for $k\ge2$, as proved by P. Erdös and J. Selfridge in doi.org/10.1215/IJM%2F1256050816 $\endgroup$
    – te4
    Commented Jan 19 at 0:07
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    $\begingroup$ Hmm, it seems to me that the definition of the function should be revisited (in OEIS begins with $1,1,2,...$), and I don't see a ratio $\log_{10}(a(n)+1)/n \to 0.8$ but a linear trend $\log_{10}(a(n)+1) \sim 0.3083 n + 0.1496$ (linear trend by Excel from 40 leading entries) $\endgroup$
    – Gottfried Helms
    Commented Jan 19 at 0:39
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    $\begingroup$ Mampo (or Mompoj) was the local indigenous chieftain (cacique) of the Malibu culture, when the Spanish conquistadors arrived, and Mompox means "land of the ruler Mampo". $\endgroup$
    – Will Jagy
    Commented Jan 19 at 4:22
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    $\begingroup$ @GottfriedHelms perhaps the $0.8$-ish was $\log_e(a_n)$ rather than $\log_{10}(a_n)$. There is also an offset issue, since obviously $a_0=a_1=1$ $\endgroup$
    – Henry
    Commented Jan 20 at 21:44

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The answer is yes. Suppose $a(m)=a(n)$ for some $m<n$. By the definition of the sequence, we have $a(n)=a(m)\prod_{i=m+1}^ni^{s_i}$, where $s_i\in\{1,-1\}$, so that the above product is equal to $1$. Letting $I_\pm=\{m<i\leq n\mid s_i=\pm 1\}$, this means $\prod_{i\in I_+}i=\prod_{i\in I_-}i$, therefore $\prod_{i=m+1}^n i=\prod_{i\in I_+}i\cdot\prod_{i\in I_-}i$ is a perfect square. This contradicts a well-known result of Erdős, which says that a product of consecutive integers cannot be a perfect square.

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    $\begingroup$ According to the Erdős and Selfridge paper linked above, this result (the one you attribute to Erdős) was also proved independently and at the same time by Rigge. $\endgroup$
    – Sam Hopkins
    Commented Jan 19 at 1:28

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