Csdrhrt

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 $f(z)=frac{1}{(1-z)(1-z^2)(1-z^3)(1-z^4)}=sum_{ngeq 0}a_n z^n$ 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 $frac{(n+5)(n^2+10n+13)}{144}.$

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 $S'={(A,B,C,D): A+B+C+D=100, Age Bge Cge D}$.

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).