MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Denote by $\omega_{d,n}(N)$ the number non-negative integer solutions of the following system of equations: \begin{gather*} \alpha_1+2 \alpha_2+\cdots+d \alpha_d=N, \end{gather*} $$ \alpha_0+\alpha_1+\alpha_2+\cdots+\alpha_d=n. $$

The following Proposition is a well-known simple combinatorial fact, see for example G.Andrews, The Theory of Partitions, Theorem 1.4.

Proposition. $\omega_{d,n}(N)=\omega_{n,d}(N).$

I am interested in a generalisation of the above in the following direction. For given integer $n,d,N_1,N_2$ consider the system of equations for a set of $\dfrac{(d+1)(d+2)}{2}$ variables $\alpha_{i,j},$ $ 0 \leq i+j \leq d :$

$$ \displaystyle \sum_{0\leq i+j \leq d}i \alpha_{i,j}=N_1, $$ $$ \displaystyle \sum_{0\leq i+j \leq d}j \alpha_{i,j}=N_2,$$ $$ \displaystyle \sum_{0\leq i+j \leq d} \alpha_{i,j}=n, $$ and denotes by $\omega_{d,n}(N_1,N_2)$ the number non-negative integer solutions of the system.

In generally $\omega_{d,n}(N_1,N_1)\neq \omega_{n,d}(N_1,N_2).$

Question. 1) For what $n,d,N_1,N_2$ we have $\omega_{d,n}(N_1,N_2)= \omega_{n,d}(N_1,N_2)?$

2) Find a recurrence relation which involves both $\omega_{d,n}(N_1,N_2)$ and $ \omega_{n,d}(N_1,N_2).$

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.