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Let $f$ be a $C^\infty$-function and define $g(x) = \exp(f(x))$.

I am interested in the higher derivatives $g^{(1)}, g^{(2)}, \ldots$ of $g$.

Let $\lambda$ be a partition of $n$, i.e. a tuple of numbers $(\lambda_1, \ldots, \lambda_k)$ satisfying $\lambda_1 + \ldots + \lambda_k = n$. We introduce the shorthand $f^{(\lambda)}$ for the product $f^{(\lambda)}(x) = f^{(\lambda_1)}(x)\cdots f^{(\lambda_k)}(x)$

It is easy to see that the derivatives of $g$ can be expressed in the form

$$g^{(n)}(x) = (\sum_{\lambda} c_\lambda f^{(\lambda)}(x))g(x)$$ where $\lambda$ runs over all partitions of $n$, e. g.:

$$g^{(2)} = (f^{(2)} + f^{(1)}f^{(1)})g$$ and $$g^{(4)} = (f^{(4)} + 4f^{(3)}f^{(1)} + 3f^{(2)}f^{(2)} + 6f^{(2)}f^{(1)}f^{(1)} + f^{(1)}f^{(1)}f^{(1)}f^{(1)} )g$$

So $c_{(2)} = c_{(1, 1)} = 1$ and $c_{(4)} = c_{(1, 1, 1, 1)} = 1$, $c_{(3, 1)} = 4$, $c_{(2, 2)} = 3$ and $c_{(2, 1, 1)} = 6$.

The question is: what are the numbers $c_\lambda$? Do they have name? Can they be easily computed from the Young tableaux? I hoped that by computing a few terms I could spot the pattern, but no luck...

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    $\begingroup$ Have you tried the online sequence finder? $\endgroup$ – aepound May 13 '15 at 14:04
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    $\begingroup$ @aepound What a cool site, didn't know about it. $\endgroup$ – Gregory Grant May 13 '15 at 14:09
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    $\begingroup$ An example of how to compute: the coefficient corresponding to the partition $4=2+1+1=2+2\cdot1$ is $$\frac{4!}{2!\,1!\,1!\cdot1!\,2!}.$$ That is, it's the multinomial coefficient $$\binom{4}{2,1,1}$$ divided by the number of permutations of parts of equal size, $1!\,2!$. Another example: the coefficient corresponding to $10=3+3+2+2=2\cdot3+2\cdot2$ is $$\frac{10!}{3!\,3!\,2!\,2!\cdot2!\,2!}.$$ In general the coefficient corresponding to a given partition of $n$ is the number of ways of partitioning a set of size $n$ into subsets of sizes given by the parts. $\endgroup$ – Will Orrick May 13 '15 at 14:26
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    $\begingroup$ You might want to have a look at Faà di Bruno's formula. $\endgroup$ – jflipp May 13 '15 at 14:29
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    $\begingroup$ Similarly to the OEIS, we created findstat.org for numbers attached to certain "combinatorial sets" (such as integer partitions). You should try plugging in your numbers above into findstat.org/StatisticFinder/IntegerPartitions. I just did, and you indeed find that they match the number of set partitions whose sorted block sizes correspond to the partition. This is exactly @WillOrrick's answer, but obtained in an automated way as the OEIS does it for sequences. $\endgroup$ – Christian May 13 '15 at 20:46

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