Q:

$s_{k+1}=\left\lfloor\frac{s_kn}{k}\right\rfloor-s_k={n-1\choose k}$? 

$$\hbox{Why }s_{k+1}=\left\lfloor\frac{s_kn}{k}\right\rfloor-s_k,\,s_0=0,\,s_1=1$$ $$\hbox{forms }{n-1\choose k-1}\hbox{?}$$
I saw it in this answer but have no explanation for this, alone to say I see it the first time while practicing math more than 15 years.  Is it at least somehow known? 
Thanks. 

A:

I don't know if this is an explanation, but $$\left\lfloor\frac{s_kn}k\right\rfloor=\left\lfloor\frac{(n-1)!n}{(k-1)!(n-k)k}\right\rfloor=\left\lfloor\binom{n}{k}\right\rfloor=\binom n k,$$
so your recursion $s_k+s_{k+1}=\lfloor\frac{s_kn}k\rfloor$ is the familiar Pascal triangle recursion for generating the binomial coefficients. 
(Composed while Luke Collins was posting his answer.)