Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How many solutions are there to $$\sum \limits_{k=0}^5a_k(6-k)=50, a_k\in\mathbb{N}_0?$$ Generalisation 1: Is there a closed form for the number of solutions to $$\sum \limits_{k=0}^na_k(n+1-k)=N, a_k\in\mathbb{N}_0?$$ Generalisation 2: Is there a general way to solve, for given $b_k\in \mathbb{N}_0$,$$\sum \limits_{k=0}^na_kb_k=N, a_k\in\mathbb{N}_0?$$

This question arose from Goldbach's 'how many ways are there to make $50$ using $7$ positive integers?'. If we let $0\le r_0\le \cdots \le r_6\le 50$, then this reduces to the number of ways of making $50$ with $6(r_0)+5(r_1-r_0)+\cdots+(r_6-r_5)$, and as all the bracketed terms are $\ge 0$ my original question results.

So far, I have considered solving $$\sum \limits_{k=0}^5c_k=50, c_k\in\mathbb{N}_0$$ then using inclusion exclusion to eliminate all cases other than when $c_k\equiv 0 \mod(6-k)$, but I doubt this will work because the floor function will appear throughout, making a closed form impossible to find by this find.

share|cite|improve this question
Maybe I'm misunderstanding your question but aren't you simply asking for the number of partitions of 50 into 7 parts? If so, there is a theory for that. – Marek Sep 25 '13 at 12:00
@Marek You're correct, thanks for the link. – Alyosha Sep 25 '13 at 18:20
For anyone interested, the answer to the final question is the coefficient of $x^{50}$ in $(1+x^{b_1}+x^{2b_1}+\cdots)(1+x^{b_2}+x^{2b_2}+\cdots)\cdots=\prod \limits_{r=1}^n\frac{1}{1-x^{b_r}}.$ – Alyosha Sep 28 '13 at 16:58
You're welcome. Consider answering the question yourself so that it's not marked as unanswered. It's even encouraged that people provide complete answers to their own questions - that way you can receive additional comments on your solution. – Marek Sep 28 '13 at 18:15
@Marek Very well, I was wary of pushing an already-answered question to the front page, and also of the fraudulent feeling of potentially receiving points for another's information. I'll probably post an answer in a short while. – Alyosha Sep 28 '13 at 19:17

If we wish to solve,

$$2x+5y=N$$ the number of solutions is the coefficient of $z^N$ in the expansion $$(1+z^2+z^{2\cdot2}+z^{3\cdot2}+\cdots)(1+z^5+z^{2\cdot 5}+z^{3\cdot 5}+\cdots)$$ as this is the number of ways of making $N$ using a positive number of $k$s and $q$s. Extending this logic to the general case

$$\sum_{i=1}^nk_ix_i=N.$$ Here the number of integer solutions is equal to the coefficient of $z^N$ in $$(1+z^{a_1}+z^{2a_1}+\cdots)(1+z^{a_2}+z^{2a_2}+\cdots)\cdots(1+z^{a_n}+z^{2a_n}+\cdots)=\prod_{i=1}^n\frac{1}{1-z^{a_i}}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.