7
votes
2answers
255 views

q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$.

The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. Consider the $q$-analog recursive definition of the ...
4
votes
2answers
282 views

Recurrence for $q$-analog for the Stirling numbers?

I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence $$ S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1) $$ with the ...