We are provided with a recurrence relation as follows:-
$F(n,k) = F(n,k-1) + F(n-k+1,k)$
$F(n,1) = n $
$F(X,k) = 0$ if $ (X\leq0)$
I need help in solving this for k=1 to 10 only
Edit:- I have added values I computed using a simple program upto n=25 and k=25( i wanted a general way for all k , but question demands only for k=1 to 10) .
I have an estimate, based on the well-known $\sum_{m=k}^n{m\choose k}={n+1\choose k+1}$. My estimate is for when $n$ is much bigger than $k$.
$$F(n,2)=F(n,1)+F(n-1,2)=F(n,1)+F(n-1,1)+F(n-2,1)+...={n+1\choose2}\\ F(n,3)=F(n,2)+F(n-2,2)+F(n-4,2)+...\approx\frac12{n+2\choose3}\\ F(n,4)\approx\frac1{3!}{n+3\choose4}$$ The fractions come because only half the F(m,2) and a third of the F(m,3) contribute to the sum. So my estimate when $n\gg k$ is $$F(n,k)\approx\frac1{(k-1)!}{n+k-1\choose k}$$
Of course, when $k>n$, then $F(n,k)=F(n,n)$