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Let $n>0$ and $s_n=\sum_{k=1}^n k$. I looked at the expressions $\displaystyle\frac{s_n!}{(s_n-n)!}$ and found that the fraction is another factorial for $k=1,2,3,4$, i.e. $$\frac{1!}{0!}=1=1!\;,\;\frac{3!}{(3-2)!}=6=3!\;,\;\frac{6!}{(6-3)!}=120=5!\;,\;\frac{10!}{(10-4)!}=5040=7! $$ The pattern stops here since $\frac{15!}{(15-5)!}= 360360\neq 9!=362880$ and I tried some more without success.

What is special about the first four? Are there other examples further out or can it be proven that they don't exist?

draks ...
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    I think it's a coincidence; the four identities don't seem to occur of the same reason. The first two are trivial, the next one nearly so, because you happen to have $3!=6$. The $6!7!=10!$ identity is remarkable coincidence but it the only identity of its kind known, and it is strongly suspected that there are no others. (See Mathworld "factorial products".) – MJD Jul 18 '13 at 21:56
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    Also relevant: On the factorial equations $A! B! =C!$ and $A!B!C! = D!$ – MJD Jul 18 '13 at 21:59
  • @MJD Do you mind posting that as an answer? – jathd Jul 18 '13 at 22:27
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    Note $$ \left(\begin{array}{c} \sum_{k=1}^{n}k\ n \end{array}\right)=\left(\begin{array}{c} \frac{1}{2}n\left(n+1\right)\ n \end{array}\right)=\frac{\left(\frac{1}{2}n^{2}+\frac{1}{2}n\right)!}{\left( \frac{1}{2}n^{2} - \frac{1}{2}n\right)!} $$ Suppose $\exists m$ s.t. $$ \left(\begin{array}{c} \sum_{k=1}^{n}k\ n \end{array}\right)=m!. $$ Then, $$ \left(\frac{1}{2}n^{2}+\frac{1}{2}n\right)!=m!\left(\frac{1}{2}n^{2}-\frac{1}{2}n\right)! $$ and either $m!$ or $\left(\frac{1}{2}n^{2}-\frac{1}{2}n\right)!$ divide $\left(\frac{1}{2}n^{2}+\frac{1}{2}n\right)!$. – parsiad Jul 18 '13 at 22:31
  • @jathd Okay, but I'm not sure it really is an answer. – MJD Jul 18 '13 at 22:42
  • @par interesting... – draks ... Jul 20 '13 at 11:32
  • @draks...: just expanding on what MJD wrote. also I made a silly mistake: should say both divide $\left(1/2 n^2 + 1/2 n\right)!$, not either or – parsiad Jul 20 '13 at 16:12

1 Answers1

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I think it's a coincidence; the four identities don't seem to occur of the same reason. The first two are trivial, the next one nearly so, because you happen to have $3!=6$. The $6!7!=10!$ identity is remarkable coincidence but it the only identity of its kind known, and it is strongly suspected that there are no others. (See the Mathworld article on "factorial products".)

So I think what you have here is the (well-known and probably unique) coincidence that $6!7! = 10$, and then that happens to line up with three smaller cases that are less surprising.

MJD
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