For a given number $n >0$ is there a way to get combination that add up to this number??
for example :
if $n=6$ then numbers that add up are $5+1,4+2,3+2+1$ so the combination is 3
if $n=4$ then numbers that add up are $3+1$ the combination is 1
i.e each number that adds up is unique and the equation don't repeat i.e $3+1$ is same as $1+3$.
what is the formula to get the such combination ??
UPDATE ok after reading few articles and finally understanding how the partition number theory works , the final question i have is given the partition of distinct numbers (which is equal to odd partiotion)
$$ \sum _{n=0}^{\infty }q(n)x^{n}=\prod _{k=1}^{\infty }(1+x^{k})=\prod _{k=1}^{\infty }{\frac {1}{1-x^{2k-1}}}. $$
how can i get the coefficient of a variable say $x^5$ of $x^{34}$
There are many formulas to calculate the partition function $p(n)$ of an integer $n$. The last was found by Professor Ken Ono in 2011. More precisely, one has, $$p(n)=\frac{\text{Tr}^{(p)}(P;n)}{24n-1}$$ The number $p(n)$ are traces of the Poincare series $P(z)$ which is defined in Prof. Ken Ono's paper. For further info, consider the following article: https://uva.theopenscholar.com/files/ken-ono/files/097.pdf