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
 A: I claim that the number of such tuples is equal to the number of partitions of 100 into 4 nonnegative integers. 
Indeed, let $S=\{(a,b,c,d): a+2b+3c+4d=100\}$, and let $S'=\{(A,B,C,D): A+B+C+D=100, A\ge B\ge C\ge D\}$. 
I claim that $S$ and $S'$ are in bijection. To see this, we will define two inverse maps between the two sets. 
Given $(a,b,c,d)\in S$, we obtain $(A,B,C,D)\in S'$ by taking
 $A=a+b+c+d$,$B=b+c+d$, $C=c+d$ and $D=d$. Conversely, given $(A,B,C,D)\in S'$, we have $(a,b,c,d)\in S$ where $d=D, c=D-C, b=B-C, a=A-B$, and it is readily verified that these are inverse assignments. 
In particular, the number of tuples you wish to count is equal to the cardinatlity of $S'$. 
Note that $S'$ is in bijection with the set of partitions of $100$ into 4 nonnegative integers. The cardinality can be expressed in terms of the partition function $p(n,k)=$ number of partitions of n into k positive integers. Indeed, the tuple (A,B,C,D) may have 0,1,2, or 3 terms equal to 0, so we have $|S'|=p(100,4)+p(100,3)+p(100,2)+p(100,1)$. The function $p(n,k)$ satisfies various recurrence relations that can be found on this page https://en.wikipedia.org/wiki/Partition_(number_theory). 
A: 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 $\frac{(n+5)(n^2+10n+13)}{144}.$

