In following, $x_{n}$ is a set of given numbers, n = 0, 1, 2, ...,
$y_{n}$ is defined by the following relation:
For example:
${\displaystyle {x_{1}=x_{0}y_{1} }}.$
${\displaystyle {x_{2}={\binom {1}{0}}x_{0}y_{2} + {\binom {1}{1}}x_{1}y_{1} }}.$
${\displaystyle {x_{3}={\binom {2}{0}}x_{0}y_{3} + {\binom {2}{1}}x_{1}y_{2} + {\binom {2}{2}}x_{2}y_{1} }}.$
For simplicity, we can assume $x_{0} = 1$.
Q: Is there an explicit solution of $y_{n}$ in term of $x_{n}$ ?
Thank you.
This is an approach of just what i meant in the comments.
Suppose $$A(z)=\sum _{i = 0}^{\infty}x_i\frac{z^i}{i!},$$ and $$B(z)=\sum _{j=1}^{\infty}jy_j\frac{z^j}{j!}$$ then $$A(z)B(z)=\sum _{n =0 }^{\infty}\frac{z^n}{n!}\left (\sum _{i =0}^{n-1}\binom{n}{i}x_i(n-i)y_{n-i}\right )=\sum _{n =0 }^{\infty}\frac{z^n}{(n-1)!}\left (\sum _{i =0}^{n-1}\binom{n-1}{i}x_i(n-i)\frac{y_{n-i}}{n-i}\right )=z\sum _{n =1 }^{\infty}\frac{z^{n-1}}{(n-1)!}\left (\sum _{i =0}^{n-1}\binom{n-1}{i}x_iy_{n-i}\right )=z\sum _{n =1 }^{\infty}\frac{z^{n-1}}{(n-1)!}x_n=zA'(z)$$ From there we get that $$B(z)=\frac{zA'(z)}{A(z)}$$ and so $$y_n=\frac{[z^{n-1}]ln(A(z))'}{n}.$$