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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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11  
There's many for example: $n!=(n!-7)+7$:) –  Sami Ben Romdhane Jul 25 '13 at 4:21
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Hmm. Given that an integral is an "infinite sum", I think $\Gamma(n+1)=n!$ infinitesimally works. –  1015 Jul 25 '13 at 4:42

10 Answers 10

up vote 40 down vote accepted

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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4  
I can't tell if you are being sarcastic - but I don't know what that sum means, can you please explain? –  zerosofthezeta Jul 25 '13 at 4:37
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@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 $n!$. –  Arkamis Jul 25 '13 at 4:38
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Nice, haha$\left . \right .$ –  Thomas Jul 25 '13 at 5:15

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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$n! = e^{ \sum_{k = 1}^n \ln (k)}$

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Thanks, but I addressed this one in my question... –  zerosofthezeta Sep 14 '13 at 7:39

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

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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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I doubt that the series converges - it is probably an asymptotic series. –  marty cohen Jul 25 '13 at 4:30
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The series "at $n=\infty$" is an asymptotic series. The series "at $n=0$" and the series "at $n=-1$" have radius of convergence $1$. –  Robert Israel Jul 25 '13 at 4:35

$$n!=\prod_{p<n}p^{\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor}$$

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Can you provide an example for a small number like 8? –  zerosofthezeta Aug 5 '13 at 3:51
    
This comes from prime factorization of $n!$. Any $p^e<n$ will divide $n!$. And $e=\sum_{k=1}^\infty\lfloor{n\over p^k}\rfloor.$ –  Kunnysan Aug 5 '13 at 4:29
    
$8! = 2^{4+2+1} 3^{2} 5^{1} 7^{1}$. –  you-sir-33433 Dec 16 '13 at 7:34

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

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