after developing a formula, I came up against this :
$$f_0(n)=2n−1,\\ f_{k+1}(n)=\sum_{i=1}^nf_k(i)$$
So, for example :$$f_1(n)= n^2\\ f_2(n) = \frac{n(n+1)(2n+1)}{6}$$ Can we find $f_n(n)$ ?
Thanks to Jens Renders for his advice of rewriting it in a more logical way
Applying the Hockey-Stick Identity repeatedly gives $$ f_1(n)=\sum_{k=1}^nf_0(k)=\binom{n}{2}+\binom{n+1}{2} $$ $$ f_2(n)=\sum_{k=1}^nf_1(k)=\binom{n+1}{3}+\binom{n+2}{3} $$ $$ f_3(n)=\sum_{k=1}^nf_2(k)=\binom{n+2}{4}+\binom{n+3}{4} $$ Following this pattern, we get $$ f_k(n)=\binom{n+k-1}{k+1}+\binom{n+k}{k+1} $$ Thus, setting $k=n$, we get $$ f_n(n)=\binom{2n-1}{n+1}+\binom{2n}{n+1} $$