Let $k, n \in \mathbb{N}$ such that we have $n > k$. My question is: In how many ordered ways can we write $n$ as sum of the natural numbers $k, k + 1, k + 2, \ldots, n$? No number less than $k$ can be used in the sum.
For example if we take $n = 10$ and $k = 2$, then we can write $10$ in $5$ ordered ways using the numbers $3,4,5,6,7,8, 9, 10$. These are
$$10 = 10 \qquad 10 = 5 + 5 \qquad 10 = 6 + 4 \qquad 10 = 7 + 3 \qquad 10 = 3 + 3 + 4$$
I have tried this for a long time and got a way to do this using the stars and bars concept, however the closed form which we get from the stars and bars seems to be very painful to evaluate for some large value of $n$ and hence I was looking for a more nice approach with a nice closed form.
Any help or hint would be highly appreciated. Thanks.
This is OEIS A026807. I doubt there is a closed form.
One way of finding this is to say $$A_{n,k}=\left\{\begin{matrix} 0 &\text{ when }&n<k \\ 1 &\text{ when }&k \le n \lt 2k \\ A_{n-k,k}+A_{n,k+1} &\text{ when }&2k \le n \end{matrix}\right.$$
since you can either have a partition of a smaller number plus the smallest extra part, or a partition of the same number with a larger minimum part, but once you are looking at a minimum part more than half the number you can only have the number itself.
So in your example, with $n=10, k=3$: $A_{10,3} $ $=A_{7,3}+A_{10,4}$ $=A_{4,3}+A_{7,4}+A_{6,4}+A_{10,5}$ $=A_{4,3}+A_{7,4}+A_{6,4}+A_{5,5}+A_{10,6}$ $=1+1+1+1+1 = 5$.