On Wikipedia's Bell polynomial Convolution identity. The convolution identity for the sequence \begin{equation} x=(x_1,x_2,...,x_n) \end{equation} and \begin{equation} y=(y_1,y_2,...,y_n) \end{equation} defined the operation $(x\diamond y)_n=\sum_{i=1}^{n-1} \binom{n}{i} x_{i} y_{n-i}$
This then defined an operation \begin{equation} x_n^{k\diamond} \end{equation} the $n$th term of the k convolution of \begin{equation} x\diamond...\diamond x \end{equation} where an example was given by \begin{equation} x=(x_1,x_2,x_3,....) \end{equation}
However, this confused me of how to obtain \begin{equation} x\diamond x =(0, 2x_1^2,6x_1x_2, 8x_1x_3+ 6x_2^2,...) \end{equation} \begin{equation} x\diamond x\diamond x =(0,0,6x_1^3, 36 x_1^2 x_2,...) \end{equation} where \begin{equation} B_{4,3}(x_1,x_2)=\frac{(x\diamond x\diamond x)_4}{3!} =6x_1^2x_2 \end{equation} because \begin{equation} (x_1,x_2)\diamond (x_1,x_2) =\binom{2}{1} x_1x_1 =2x_1^2 \end{equation} I don't see where did the $0$ in $x\diamond x$ or $6x_1x_2$ come from.
What's the correct definition for this expression?
We have \begin{align*} (x\diamond y)_n&=\sum_{j=1}^{n-1}\binom{n}{j}x_jy_{n-j}\qquad\qquad n\geq 1\\ (x\diamond y)&=\left((x\diamond y)_1,(x\diamond y)_2,(x\diamond y)_3,\ldots\right)\\ &=\left((x\diamond y)_n\right)_{n\geq 1} \end{align*} We obtain according to this definition \begin{align*} \color{blue}{(x\diamond x)_1}&=\sum_{j=1}^{0}\binom{1}{j}x_jx_{1-j}\color{blue}{=0}\tag{$\to$ empty sum}\\ \color{blue}{(x\diamond x)_2}&=\sum_{j=1}^1\binom{2}{j}x_jx_{2-j}=\binom{2}{1}x_1x_1\color{blue}{=2x_1^2}\\ \color{blue}{(x\diamond x)_3}&=\sum_{j=1}^2\binom{3}{j}x_jx_{3-j}\\ &=\binom{3}{1}x_1x_2+\binom{3}{2}x_2x_1\color{blue}{=6x_1x_2}\\ \color{blue}{(x\diamond x)_4}&=\sum_{j=1}^3\binom{4}{j}x_jx_{4-j}\\ &=\binom{4}{1}x_1x_3+\binom{4}{2}x_2x_2+\binom{4}{3}x_3x_1\\ &\color{blue}{=8x_1x_3+6x_2^2} \end{align*} It follows \begin{align*} \color{blue}{(x\diamond x)=(0,2x_1^2,6x_1x_2,8x_1x_3+6x_2^2,\ldots)} \end{align*} The expression \begin{align*} (x\diamond x\diamond x)=((x\diamond x)\diamond x)=(x\diamond (x\diamond x)) \end{align*} can be calculated analogously.