Let a sequence be defined recursively ($y_0=1$):
$$y_n=\sum_{j=0}^{n-1}\binom{n-1}{j}y_jc_{n-j}=\sum_{j=0}^{n-1}\frac{(n-1)!}{j!(n-1-j)!}y_jc_{n-j}$$
I would like the recursive definition for $z_n = y_n/\sqrt{n!}$. However, this doesn't work:
$$z_n=\frac{y_n}{\sqrt{n!}}\neq\sum_{j=0}^{n-1}\sqrt{\frac{n!}{j!}}\frac{1}{n(n-1-j)!}z_jc_{n-j}$$
Why? What's the correct way to obtain a recursive definition for $z_n$? How about for $w_n = y_n/n!$?
$$\sqrt{n!}z_n=y_n=\sum_{j=0}^{n-1}\frac{(n-1)!}{j!(n-1-j)!}y_jc_{n-j}=\sum_{j=0}^{n-1}\frac{(n-1)!}{j!(n-1-j)!}\sqrt{j!}z_jc_{n-j},$$ so after dividing by $\sqrt{n!}$, $$z_n=\frac1{\sqrt n}\sum_{j=0}^{n-1}\frac{\sqrt{(n-1)!}}{\sqrt{j!}(n-1-j)!}z_jc_{n-j}.$$
$${n!}w_n=y_n=\sum_{j=0}^{n-1}\frac{(n-1)!}{j!(n-1-j)!}y_jc_{n-j}=\sum_{j=0}^{n-1}\frac{(n-1)!}{j!(n-1-j)!}{j!}w_jc_{n-j},$$ so after dividing by ${n!}$, $$w_n=\frac1{ n}\sum_{j=0}^{n-1}\frac{1}{(n-1-j)!}w_jc_{n-j}.$$