# Is there an explicit solution of $y_{n}$ in this binomial coefficient relation?

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.

• Being off by an index it is the binomial transform. Nov 16, 2019 at 14:26
• @Phicar, can you explain a bit more on your comment ? I could not see why original relation is a binomial transform. Nov 22, 2019 at 5:44

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 $$y_n=\frac{[z^{n-1}]ln(A(z))'}{n}.$$
• Yes, binomial transform is just multiplying by exponential in the exponential generating function. If you know your sequence $x_i$ perhaps you can express formally its exponential function and compose. Perhaps you will find interesting the book "Combinatorial Species and Tree-like Structures". Usually logarithm corresponds to understand your sequence as parts of a finer sequence. Take a look, for example, to Stirling numbers of the first kind. Nov 22, 2019 at 19:46
• It is still not clear to me, do $x_{n}, y_{n}$ or $x_{n}/n! , y_{n+1}/n!$ in the expression ${\displaystyle {x_{n}=\sum _{i = 0}^{n - 1}{\binom {n - 1}{i}}x_{i}y_{n -i} }}.$ form a binomial transform pair ? Nov 23, 2019 at 1:18