# Explicit form of a recursive relation

My question is to find an explicit formula for $$N_{k,n}$$, if the following recursive one is known:

\left\{ \begin{align} \begin{split} N_{1,n} &= 1 \text{,} \\ N_{k+1,n} &= \sum_{i = k+1}^{n-1} N_{k,i} \end{split} \end{align} \right. $$\text{for } n \in \{\,2, 3, \dots\,\}, k \in \{\,n+1, n+2, \dots\,\} \text{.}$$

I can calculate a few first terms ($$N_{1,n} = 1$$, $$N_{2,n} = n-2$$, $$N_{3,n} = \frac{(n-2)(n-3)}{2}$$), but unfortunately I cannot find the general solution.

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## 1 Answer

As mentioned in comments, one way is to notice the Hockey stick identity in the definition and then prove it matches for rest of your definition, giving the closed form $$N_{k,n}=\binom{n-2}{k-1}$$.

If you don't notice that at first, you can simplify the recurrence $$N_{k+1,n+1} - N_{k+1,n} = \sum_{i=k+1}^{n}N_{k,i}-\sum_{i=k+1}^{n-1}N_{k,i} = N_{k,n},$$ hence $$N_{k+1,n+1} = N_{k+1,n} + N_{k,n}.$$ Now add base cases such as $$N_{1,n}=1$$ and $$N_{k,2}=0$$ for $$k\geq 2$$, which follow trivially from definitions. Again you can notice that the recurrence definition resembles Pascal's formula and you can go from there.

Eventually if all that fails, you can use a generic method for solving the last recurrence. For example, using generating functions as in the answer Solving two-dimensional recurrence relation $a_{i,\ j}\ =\ a_{i,\ j-1}\ +\ a_{i-1,\ j-1}$, you should get $$f(x,y)=\sum_{k\geq 1, n \geq 2}N_{k,n}x^k y^n=\frac{xy^2}{1-y-xy}$$. This in turn simplifies using geometric series into $$f(x,y)=xy^2\sum_{n=0}^{\infty}y^n(1+x)^n=xy^2\sum_{n=0}^{\infty}y^n\sum_{k=0}^{n}\binom{n}{k}x^k = \sum_{n\geq 0, k \geq 0}^{\infty}\binom{n}{k}x^{k+1}y^{n+2},$$ and so finally $$f(x,y)=\sum_{n\geq 2, k \geq 1}^{\infty}\binom{n-2}{k-1}x^{k}y^{n}.$$ Now just read off the coefficient of $$x^k y^n$$.