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$f(k)={k \choose k-1} f(k-1) + { k \choose k-2} f(k-2) + .... {k \choose 3} f(3)$

$f(3) = 1$

$k \ge 3$

Even good upper and lower bounds will help me as I am trying to find how this function grows asymptotically.

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This is quite similar to the recurrence for Bell Numbers: – Aryabhata Jun 5 '11 at 8:23
up vote 3 down vote accepted

Consider the exponential generating function $$ F(s)=\displaystyle\sum_{k\ge3}f(k)\frac{s^k}{k!}. $$ Your recursion on the coefficients $(f(k))$ can be translated into a functional equation on $F$. Unless I am mistaken, one gets $$ F(s)=\frac{s^3}{6(2-\mathrm{e}^s)}, $$ hence the first singularity is at $s=\log2$. From there, the asymptotics on $f(k)$ when $k\to+\infty$ is $$ f(k)=k^k(\log 2)^k\mathrm{e}^{-k+o(k)}. $$

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