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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.

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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
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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
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1 Answer 1

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 $$(1+z^{a_1}+z^{2a_1}+\cdots)(1+z^{a_2}+z^{2a_2}+\cdots)\cdots(1+z^{a_n}+z^{2a_n}+\cdots)=\prod_{i=1}^n\frac{1}{1-z^{a_i}}.$$

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