Relation between stirling numbers

Question

Is there a relation between $$ \genfrac\{\}{0pt}{}{n}{n-2} $$

and $$ \genfrac\{\}{0pt}{}{n-1}{n-3} $$

Like the first one can be obtained from the second one by adding something?

Answer 1

The Stirling numbers of the second kind satisfy a somewhat Pascal-like recurrence relation:

$${{n+1}\brace k}=k{n\brace k}+{n\brace{k-1}}\;.$$

In particular, then,

$${n\brace{n-2}}=(n-2){{n-1}\brace{n-2}}+{{n-1}\brace{n-3}}\;.\tag{1}$$

It’s easy to check that

$${{n-1}\brace{n-2}}=\binom{n-1}2\;,$$

so $(1)$ becomes

$${n\brace{n-2}}=(n-2)\binom{n-1}2+{{n-1}\brace{n-3}}\;.$$