Show that an explicit formula for the partition transform coefficients $C$ OEIS A134685 for the extended Lagrange Inversion exist, and that such formula is the following one $$C_{u,m}=(-1)^{m_1+m_2+\dots+m_{n}}\frac{1\times 2\times 3\times\dots\times(n+m_1+\dots+m_{n})}{(u_1+1)!(u_2+1)!\dots(u_{n}+1)!m_1!m_2!\dots m_{n-1}!}.$$
Basically this result allows us to make explicit the $n$th derivative of any inverse function $ g $ of $ f $, at the point $x_0 = f (y_0)$, if the derivatives of the function $ f $ are known
I would like to share this result to find out if it is correct, show it to those who may be interested and implement the topic thanks to any answers that go to fill / correct any my mistakes
Let $ f $ be an analytic function and let $ g $ be its inverse, both expressible as a power series, denoting by $ g ^ {\{n \}} (x) $ the $ n $ -th derivative of $ g $ valued in $ x $:
$$ g(x)=g(x_0)+\sum_{n=1}^\infty \frac{g^{\{n\}}(x_0)}{n!}(x-x_0)^n \tag{1}$$
$$ f(g(x))=x=f\big( g(x_0)\big)+\sum_{n=1}^\infty \frac{f^{\{n\}}\big(g(x_0)\big)}{n!}(g(x)-g(x_0))^n \tag{2}$$
Lagrange's inversion theorem allows us to express the inverse function $ g $ exclusively in terms of derivatives of $ f $, denoting the derivative $ n $ -th of $ g $ as: $$\ g^{\{ n\}}(x_0)=\lim_{x\to x_0}\frac{d^{n-1}}{d g_{(x)}^{n-1}} \Bigg[\Big( \frac{g(x)-g(x_0)}{f\big(g(x)\big)-f\big(g(x_0)\big)}\Big)^n\Bigg] \tag{3}$$ To find $ g ^ {\{n \}} (x_0) $ we must derive $ n-1 $ times the term in square brackets, using the function $ g (x) $ as the derivation variable. Then we should take the limit for $ x \to x_0 $ or equivalently, by setting $ y = g (x) -g (x_0) $, make it the limit for $ y \to 0 $
$$ g^{\{ n\}}(x_0)=\frac{d^{n-1}}{d y^{n-1}}\Bigg[ \frac{1}{\Big(f^{\{1\}}\big(g(x_0)\big)+f^{\{2\}}\big(g(x_0)\big)y+f^{\{3\}}\big(g(x_0)\big)y^2+\dots\Big)^{n}}\Bigg]_{y=0} \tag{4} $$
$ \ $
We call $ a (y): = 1 / y $ e $ b (y): = \big (f ^ {\{1 \}} (g) + f ^ {\{2 \}} (g) y + f ^ {\{3 \}} (g) y ^ 2 + \dots \big) ^ {n} $ and we apply Faà di Bruno's formula to $ (a \circ b) (y) $. Let $ m_1, m_2, \dots, m_n $ be non-negative integers that satisfy the equation $ 1m_1 + 2m_2 + \dots + v \ m_v = v $ and let $ k = m_1 + m_2 + \dots + m_v $ $$(a \circ b)^{\{ v\}}(y)=\sum \frac{v!}{m_1! m_2! \dots m_v!} a^{\{k\}}\big(b(y)\big) \Big[\frac{b^{\{1\}}(x)}{1!}\Big]^{m_1} \Big[\frac{b^{\{2\}}(x)}{2!}\Big]^{m_2}\dots\Big[\frac{b^{\{n\}}(x)}{n!}\Big]^{m_v} \tag{5}$$
$$ a^{\{ k\}}(z)=(-1)^k\frac{k!}{z^{k+1}}\Rightarrow a^{\{ k\}}(b(y))=(-1)^k\frac{k!}{b(y)^{k+1}} \tag{6}$$ From the multinomial theorem we have that $ j_1 + j_2 + \dots = n $ $$b(y)=\big(f^{\{1\}}(g)+f^{\{2\}}(g)y+\dots\big)^{n}=\sum \frac{n!}{j_1!j_2!\dots}\Big[ f^{\{1\}}(g)\Big]^{j_1}\Big[ f^{\{2\}}(g)y\Big]^{j_2}\dots \tag{7}$$ From Leibniz's general rule of product of derivatives, we have that $ u_1 + u_2 + u_3 \dots = w $
$$\frac{d^w}{d y^w}[ f^{\{1\}}(g)]^{j_1}\times [ f^{\{2\}}(g)y]^{j_2}\times \dots=\sum \frac{w!}{u_1! u_2! \dots} \frac{d^{u_1}}{d y^{u_1}} \Big\{\big[ f^{\{1\}}(g)\big]^{j_1}\Big\}\frac{d^{u_2}}{d y^{u_2}} \Big\{\big[ f^{\{2\}}(g)y\big]^{j_2}\Big\}\dots\tag{8}$$ Since we will have to evaluate $ (a \circ b) ^ {\{v \}} (y) $ with $ y = 0 $ we see that by substituting this value in the equation $ (8) $, the only terms of the summation that will not be zero will be those for which $ u_i = (i-1) j_i $. Combining the equations $ (4), (5), (6), (7), (8) $ we finally get:
$$C_{u,m}=(-1)^{m_1+m_2+\dots+m_{n}}\frac{1\times 2\times 3\times\dots\times(n+m_1+\dots+m_{n})}{(u_1+1)!(u_2+1)!\dots(u_{n}+1)!m_1!m_2!\dots m_{n-1}!}\tag{9}$$
$$ g^{\{n+1\}}(x)=\sum \frac{C_{u,m}}{f'(g_{(x)})^{n+1}} \frac{f^{\{u_1+1\}}(g_{(x)})}{f'(g_{(x)})} \frac{f^{\{u_2+1\}}(g_{(x)})}{f'(g_{(x)})}\dots\frac{f^{\{u_n+1\}}(g_{(x)})}{f'(g_{(x)})}$$ $x=f(y)$ $$ g^{\{n+1\}}(x)=\sum \frac{C_{u,m}}{f'(y)^{n+1}} \frac{f^{\{u_1+1\}}(y)}{f'(y)} \frac{f^{\{u_2+1\}}(y)}{f'(y)}\dots\frac{f^{\{u_n+1\}}(y)}{f'(y)}$$ With $ m_1, m_2, \dots, m_n $, $ u_1, u_2, \dots, u_n $ non-negative integers that satisfies the equations:
$$ u_1 + u_2+ \dots u_n= n \tag{10}$$ $$ u_{a_1}=u_{a_2}=u_{a_3} =\dots= u_{a_j}=k \Leftrightarrow j=m_k $$ $$ 1 m_1+2m_2+\dots +k m_k+\dots+ n m_n=n \tag{11}$$ If there are $ j $ integers $ u $ in $ (10) $ that have the same value $ k $, in $ (11) $ you will have $ m_k = j $. In other words, if $ \mathbf {u} = (u_1, u_2, \dots, u_n) $ a solution of the equation $ (10) $, the $ k $ -th $ m_k $ component of the vector $ \mathbf { m} = (m_1, m_2, \dots, m_n) $ solution of $ (11) $ represents the frequency of the non-zero integer $ k $ in the vector $ \mathbf {u} $. The sum of the coefficients of the vector $ \mathbf {m} $ will return the number of non-zero coefficients of the vector $ \mathbf {u} $.
Example1: Let $ n = 10 $, and $ \mathbf {u} = (3,2,2,1,1,1,0,0,0,0) $ a solution of $ (10) $, since the number "1" appears three times in $ \mathbf {u} $, "2" twice, "3" once and all other zero times, the associated solution of $ (11) $ that will appear simultaneously under the sum, will be $ \mathbf {m} = (3,2,1,0,0,0,0,0,0,0) $.
Example2: if we want to calculate the coefficients of $g^{\{6\}}(x)$, so we take the integer partitions of $6-1=5$: $$\mathbf {u}_1=(5);\mathbf {u}_2=(1,4);\mathbf {u}_3=(2,3);\mathbf {u}_4=(1,1,3);\mathbf {u}_5=(1,2,2);\mathbf {u}_6=(1,1,1,2);\mathbf {u}_7=(1,1,1,1,1) $$ $$\mathbf {m}_1=(1);\mathbf {m}_2=(1,1);\mathbf {m}_3=(1,1);\mathbf {m}_4=(2,1);\mathbf {m}_5=(1,2);\mathbf {m}_6=(3,1);\mathbf {m}_7=(5) $$
$$C_{\mathbf {u}_1,\mathbf {m}_1}=(-1)^{1}\frac{1\times 2\times \dots\times(5+1)}{(5+1)! \times 1!}=-\frac{6!}{6!}=-1$$ $$C_{\mathbf {u}_2,\mathbf {m}_2}=(-1)^{1+1}\frac{1\times 2\times \dots\times(5+1+1)}{(1+1)!(4+1)! \times 1!1!}=\frac{7!}{2!5!}=21$$ $$C_{\mathbf {u}_3,\mathbf {m}_3}=(-1)^{1+1}\frac{1\times 2\times \dots\times(5+1+1)}{(2+1)!(3+1)! \times 1!1!}=\frac{7!}{3!4!}=35$$ $$C_{\mathbf {u}_4,\mathbf {m}_4}=(-1)^{2+1}\frac{1\times 2\times \dots\times(5+2+1)}{(1+1)!(1+1)!(3+1)! \times 2!1!}=\frac{8!}{2!2!2!4!}=-280$$ $$C_{\mathbf {u}_5,\mathbf {m}_5}=-210; \ \ C_{\mathbf {u}_6,\mathbf {m}_6}=1260; \ \ C_{\mathbf {u}_7,\mathbf {m}_7}=-945$$
Writing the $ k $ -th derivative $ f ^ {\{k \}} (g (x)) $ as $ f_ {k} $ while the apex refer to the multiplicative power
$$ g^{\{6\}}(x)f_{1}^{11} =-945f^5_2+1260f_1f_2^2f_3-210f_1^2f_2^2f_4-280f_1^2f_2f_3^2+21f_1^3f_2f_5+35f_1^3f_3f_4-f_1^4f_6 $$ We define the function InverseD[f, g][x, n] in Mathematica language which returns the $ n $ -th derivative in $ x $ of the inverse $ g $ function of $ f $ in terms of the coefficients CC[n,i]:
P[n_]:=PartitionsP[n];
u[n_]:=IntegerPartitions[n];
m[n_]:=Table[Table[Count[u[n][[j]],k],{k,1,n}],{j,1,P[n]}];
CC[n_,i_]:=(-1)^ (Total@m[n][[i]])Total@(u[n][[i]]+1)!/(Times@@((u[n][[i]]+1)!)Times@@(m[n][[i]]!));
InverseD[f,g]:=1/f'[g[#1]]^ (#2) Sum[CC[#2-1,i]Product[Derivative[1+u[#2-1][[i,j]]][f][g[#1]]/f'[g[#1]], {j,1,Total@m[#2-1][[i]]}],{i,1,P[#2-1]}]&;