Ways to express a number with a sum of factorials of $n \geq 2$ I am wondering how I should express a number with a sum of factorials. I know all numbers can be expressed with $1!$ obviously, but how should I go about expressing a number as a sum of factorials $\geq 2$. For example if we take $ 10$, it can be expressed as $3!+2!+2!, 2!+2!+2!+2!+2!$
or $12 = 3!+3!, 3!+2!+2!+2!, 6 \cdot 2!$
I understand that odd numbers can't be expressed in this way at all, however what would be the technique for finding all ways to express even numbers as factorials? 
Thanks! 
 A: a) the "compact" representation
Same as what you do in representing a number in base $10$ or $2$, etc.
you can do for representing it in the "factorial" base $1,2, 6, \cdots, n!$: refer to this Wikipedia article.
Only, instead of having a fixed ratio between the terms of the base (e.g. $10$, in the decimal)
you will have a variable ratio  $= n$, but that does not affect the algorithm substantially.
So let's take for instance $31=3 \cdot 10^1 + 1 \cdot 10^0= 10^1+10^1+10^1+10^0$.
$4! \le 31 < 5!$ so in the factorial base we will have
$$
\eqalign{
  & 31 = \left\lfloor {{{31} \over {4!}}} \right\rfloor 4! + 31\bmod 4! = 1 \cdot 4! + 7  \cr 
  & 7 = \left\lfloor {{7 \over {3!}}} \right\rfloor 3! + 7\bmod 3! = 1 \cdot 3! + 1  \cr 
  & 1 = \left\lfloor {{1 \over {2!}}} \right\rfloor 2! + 1\bmod 2! = 0 \cdot 2! + 1  \cr 
  & 1 = \left\lfloor {{1 \over {1!}}} \right\rfloor 1! + 1\bmod 1! = 1 \cdot 1! + 0  \cr 
  & \quad  \Downarrow   \cr 
  & 31 = 4! + 3! + 1! \cr} 
$$
b) how many representations
Understanding that you are asking in how many ways we can represent $x$
as a linear combination of factorials
$$
x = t_2 2! + t_3 3! +  \cdots  + t_n n!
$$
that is how many $n$-uples $(t_2  , t_3  ,  \cdots  , t_n)$ can we find
such that the above identity is satisfied, given that $n$ is the max 
for which $n! \le x$.
Starting from the "compact" representation
$$
x = c_2 2! + c_3 3! +  \cdots  + c_n n!
$$
you can decide to downgrade  either none, or one, or two, .., or or all the $c_n$ terms in $n!$
down to $n \cdot (n-1)!$.
The choices you have are 
$$
\sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {1 }  = c_{n } + 1
$$
That will make $c_{n-1}$ to become $c_{n-1},\;c_{n-1}+n,\;c_{n-1}+2n,\;\cdots,c_{n-1}+c_{n}n,\;$.
Then in turn, for each of the $c_{n-1}+k_{n} n$ you can decide to downgrade none, or one, or two..
Now, the total of choices becomes
$$
\eqalign{
  & \sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {\sum\limits_{0\, \le \,k_{\,n - 1} \, \le \,c_{\,n - 1}  + k_{\,n} n} 1 }  = \sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {\,c_{\,n - 1}  + 1 + k_{\,n} n}  =   \cr 
  &  = \left( {c_{\,n - 1}  + 1} \right)\left( {c_n  + 1} \right) + \left( \matrix{
  c_n  + 1 \cr 
  2 \cr}  \right)n \cr} 
$$
At the next step we will have 
$$
\eqalign{
  & \sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {\sum\limits_{0\, \le \,k_{\,n - 1} \, \le \,c_{\,n - 1}  + k_{\,n} n} {\sum\limits_{0\, \le \,k_{\,n - 2} \, \le \,c_{\,n - 2}  + k_{\,n - 1} \left( {n - 1} \right)} 1 } }  =   \cr 
  &  = \sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {\sum\limits_{0\, \le \,k_{\,n - 1} \, \le \,c_{\,n - 1}  + k_{\,n} n} {c_{\,n - 2}  + 1 + k_{\,n - 1} \left( {n - 1} \right)} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k_{\,n} \, \le \,c_{\,n} } {\left( {c_{\,n - 2}  + 1} \right)\left( \matrix{
  c_{\,n - 1}  + k_{\,n} n + 1 \cr 
  1 \cr}  \right) + \left( {n - 1} \right)\left( \matrix{
  c_{\,n - 1}  + k_{\,n} n + 1 \cr 
  2 \cr}  \right)}  \cr} 
$$
which does not look that may lead to a closed form.
