counting the number of non-negative integer pairs of $a+2b+3c+4d=100$ I thought partition number $P(n,4)$ is equal with 
the number of non-negative integer pairs $a+2b+3c+4d=n-4$ 
so....
how do I count the number of integer pairs satisfying $a+2b+3c+4d=100$ over $a,b,c,d \geq 0$ ??
I really wonder this problem!
thank you for sharing your knowledge.
please don't use a computer

Expanding my comment: if we consider the partial fraction decomposition of $f(z)=\frac{1}{(1-z)(1-z^2)(1-z^3)(1-z^4)}=\sum_{n\geq 0}a_n z^n$ we have that the pole at $z=1$ contributes with
$$ \frac{1}{24(1-z)^4}+\frac{1}{8(1-z)^3}+\frac{59}{288(1-z)^2}+\frac{17}{72(1-z)} $$
to $f(z)$, hence it contributes with
$$ \frac{2n^3+30n^2+135n+175}{288}$$
to $a_n$. The pole at $z=-1$ contributes with
$$ (-1)^n \frac{n+5}{32} $$
and the other poles provide bounded contributions, since they are simple. It follows that for even values of $n$
$$ a_n \approx \frac{(n+5)(n^2+10n+22)}{144} $$
and for $n=100$ the ceiling of the RHS is exactly $a_{100}$, i.e. $8037$. The error of this approximation is bounded by twice the sum of the absolute values of the real parts of the residues at the simple poles, i.e. $\frac{2}{9}<\frac{1}{2}$. In particular 

If $n$ is even $a_n=[x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)}$ is the closest integer to $\frac{(n+5)(n^2+10n+22)}{144}.$
  If $n$ is odd $a_n=[x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)(1-x^4)}$ is the closest integer to $\frac{(n+5)(n^2+10n+13)}{144}.$

