2
$\begingroup$

It is well know that for fixed $k$ the asymptotic approximation for the Stirling numbers of the second kind is given by $\frac{k^n}{k!}$. Does such simple asymptotic expression also exist for the r-associated Stirling numbers of the second kind? Thank you,

$\endgroup$
0
$\begingroup$

From the recurrence identity $S_2(n,k)=n! \sum_{j=0}^k \frac{(-1)^jS(n-j-k-j)}{j!(n-j)!}$ given by Howard and the approximation for the Stirling number of the second kind we can get that:

$S_2(n,k) \sim \frac{k^n}{k!}$ and maybe more interesting that $S(n,k)-S_2(n,k) \sim \frac{n(k-1)^{n-1}}{(k-1)!}$

Higher order associated Stirling of the second kind can be derived in a similar way by analyzing the general recurrence identity given at Howard.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.