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I know of the sum of the natural logarithms of the factors of n! , but would like to know if any others exist.

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    $\begingroup$ There's many for example: $n!=(n!-7)+7$:) $\endgroup$
    – user63181
    Jul 25, 2013 at 4:21
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    $\begingroup$ Hmm. Given that an integral is an "infinite sum", I think $\Gamma(n+1)=n!$ infinitesimally works. $\endgroup$
    – Julien
    Jul 25, 2013 at 4:42

13 Answers 13

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This one is pretty important: $$n! = \sum_{\sigma\in S_n} 1$$

Edit: As Arkamis explains, $S_n$ is the symmetric group on $n$ letters. Each $\sigma\in S_n$ is a permutation on the set $[1,2,\ldots,n]$. Since $S_n$ is a finite set, we may sum a function over it, and the sum of the constant function $f(\sigma)=1$ is just the size of the set, which is $|S_n| = n!$.

Arguably, summing a constant function is cheating. Here's one way to raise the stakes. Let $B_n$ be the set of $n\times n$ integer matrices $A$ such that every sum of a subset of entries from $A$ is in $[0,n]$. Then:

$$n!=\sum_{A\in B_n}|\det A|$$

This is the same identity in a more interesting disguise. Every $n\times n$ permutation matrix $A$ is a member of $B_n$, and $\det A = \pm 1$. On the other hand, if $A\in B_n$ is not a permutation matrix, then you can prove that $\det A = 0$.

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    $\begingroup$ I can't tell if you are being sarcastic - but I don't know what that sum means, can you please explain? $\endgroup$ Jul 25, 2013 at 4:37
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    $\begingroup$ @Euclid It means that the order of the symmetric group $S_n$ is $n!$. $\sigma \in S_n$ means "each element $\sigma$ in the group $S_n$, of which there are $n!$. $\endgroup$
    – Emily
    Jul 25, 2013 at 4:38
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Lots of good answers, but the answer is quite easily, "yes."

Integer multiplication is repeated addition.

$n!$ is $n$ groups of $(n-1)!$ objects, so $n! = \sum_{k=1}^n (n-1)!$.

Then, $(n-1)!$ is $n-1$ groups of $(n-2)!$ objects, so $n! = \sum_{k_1=1}^n \sum_{k_2 = 1}^{n-1} (n-2)!$, and so on.


Example:

$$4! = \sum_{k_1 = 1}^4 3! = 3!+3!+3!+3! = 4\cdot 3!.$$ $$\begin{align*} 4! &= \sum_{k_1=1}^4 \sum_{k_2=1}^3 2! \\ &= \sum_{k_1=1}^4 2!+2!+2! \\ &= 2!+2!+2! + 2!+2!+2! + 2!+2!+2! + 2!+2!+2! \\ &= 12\cdot 2! \\ &=4\cdot 3\cdot 2!. \end{align*}$$

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$$n! = 1 + \sum_{k=1}^n (k-1) (k-1)!$$

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  • $\begingroup$ Can you explain this please? $\endgroup$
    – Gautam
    Sep 4, 2017 at 6:53
  • $\begingroup$ What part about it don't you understand? $\endgroup$ Sep 4, 2017 at 7:04
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$n! = e^{ \sum_{k = 1}^n \ln (k)}$

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    $\begingroup$ Thanks, but I addressed this one in my question... $\endgroup$ Sep 14, 2013 at 7:39
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Sure, for $n\ge 1$ we have $$n!=\fbox{$(n-1)\times (n-1)!$} + \fbox{$(n-1)!$}$$

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If you really want, we can write $n!$ using this infinite series

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    $\begingroup$ I doubt that the series converges - it is probably an asymptotic series. $\endgroup$ Jul 25, 2013 at 4:30
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    $\begingroup$ The series "at $n=\infty$" is an asymptotic series. The series "at $n=0$" and the series "at $n=-1$" have radius of convergence $1$. $\endgroup$ Jul 25, 2013 at 4:35
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$$n!=\prod_{p<n}p^{\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor}$$

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    $\begingroup$ Can you provide an example for a small number like 8? $\endgroup$ Aug 5, 2013 at 3:51
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    $\begingroup$ This comes from prime factorization of $n!$. Any $p^e<n$ will divide $n!$. And $e=\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor.$ $\endgroup$
    – Kunnysan
    Aug 5, 2013 at 4:29
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    $\begingroup$ $8! = 2^{4+2+1} 3^{2} 5^{1} 7^{1}$. $\endgroup$
    – Slade
    Dec 16, 2013 at 7:34
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Another combinatorially important sum: $\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$, where $\displaystyle\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ is the (unsigned) Stirling symbol of the first kind, which can be defined (for instance) by the relation $x(x+1)\ldots(x+n-1) = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr] x^k$. (And of course, by setting $x=1$ in this relation we get the initial sum.) The combinatorial significance is that $\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ counts the number of permutations of the numbers $1..n$ with $k$ distinct cycles; the relation $\displaystyle n! = \sum_{k=0}^n\bigl[\begin{smallmatrix}n\\k \end{smallmatrix}\bigr]$ thus says that the total number of permutations is just the sum over all $k$ of the number of permutations into $k$ cycles.

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I have found the following formula: $$n!=\sum_{k=2}^{n+1}(-1)^{n+1-k}\binom{n+1}{k}\sum_{i=1}^{k-1}i^{n},\ n\in N.$$

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I know this question is old, but I found it googling for summation definitions of the factorial function, and thought a good answer would be the following formula, which I discovered several years ago and I haven't seen anywhere else:

$n!=\displaystyle\sum_{k=1}^n \binom{n}{k} (-1)^{n+k} k^n$

I never took the time to find a proof, but it certainly works. It would be interesting if anyone with decent math skills (that is, better than me) could shed some light on it.

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$$\frac{(n+2)!}{2} = (n+1)! + \sum_{i=1}^n i\,n!$$

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$$(v+1)! = 1 + \sum_{k=1}^v (v-k-1)! (v-k)^2$$

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    $\begingroup$ Explaining why this formula is true would be a good improvement to this answer. $\endgroup$
    – apnorton
    Jan 27, 2015 at 20:55
  • $\begingroup$ Sorry i have dyslexia ,change the left side of the equation with ν! $\endgroup$
    – kostas
    Jan 28, 2015 at 3:52
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Here's a good reference document from a few years back: Factorials as sums by Roberto Anglani and Margherita Barile.

In this paper we give an additive representation of the factorial, which can be proven by a simple quick analytical argument. We also present some generalizations, which are linked, on the one hand to an arithmetical theorem proven by Euler (decomposition of primes as the sum of two squares), and, on the other hand, to modern combinatorics (Stirling numbers).

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