Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

What is the closed form solution to the following partial recurrence relation? $$f(k,n) = \sum\limits_{t=0}^{m}f(k-t, n-1),$$ where $ m \geq 0$ is some fixed parameter.

The boundary values are $f(k,n)=0$ for all $k < 0, n < 1$ except $f(0,0)=1$.

I noticed that for $m \geq k$ its closed form is a binomial coefficient $f(k,n)=C_{k+n-1}^{n-1}$, $k \geq 0$, $n > 0$. Do you have any idea how to solve this equation for $m<k$? (Should be something like $f(k,n)=C_{k+n-1}^{n-1} - A(k,n)$).

share|cite|improve this question

If $m < k$ and $k > 0$, then your sum is adding zeroes together, since by applying your recurrence rule once on each term of the sum, you hit on negative parameters, for which the value of $f$ is zero.

So for $m < k$ :

  • $k > 0 \implies f(k,n) = 0$
  • $k = 0 \implies m < 0$ so your recurrence is undefined (or equal to $0$)
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.