1. finding an explicit formula for the coefficients of the composition of generating functions and their degrees $$ F(x,y,\ldots,G(x,y,\ldots,z),\ldots,z)^k $$
2. finding explicit formulas for the mutual generating function and their degrees $$ \frac{1}{G(x,y,\ldots,z)^k} $$
3. finding explicit formulas for a generating function of the form $$ G(x,y,\ldots,z)^{\alpha} , \alpha\in R $$
4. finding explicit formulas for the inverse generating function and their degrees; $$ G(x,y,\ldots,z)=\frac{y}{F\left(x,y\cdot G(x,y,\ldots,z),\ldots,z\right)} $$
5. finding explicit formulas for a reversible generating function and their degrees; \ $$ G(x,y,\ldots,z)=F(x,y\,G(x,y,\ldots,z),\ldots,z) $$
6. finding explicit formulas for a reversible generating function and their degrees;\ $$ G(x,y,\ldots,z)=F(x,y\,G(x,y,\ldots,z),\ldots,z) $$
7. finding explicit formulas for the inverse inverse generating function of two variables and their degrees; $$ G(x,y,\ldots,z)=F\left(x,\frac{y}{G(x,y,\ldots,z)},\ldots,z\right) $$
8. finding explicit expressions of logarithmic derivatives $$ \log(G(x,y,\ldots,z)) $$

• generating function
• ordinary generating function
• exponential generating function
• generating function of one variable
• generating function of two variables
• generating function of many variables
• coefficient of the generating function
• coefficient of the generating function of one variable
• coefficient of the generating function of two variables
• coefficient of the generating function of many variables
• coefficient of the kth degree of the generating function
• coefficient of the kth degree of the generating function of one variable
• coefficient of the kth degree of the generating function functions of two variables
• coefficient of the kth degree of the generating function of many variables
• of the composite
• functional equation
• composition of generating functions
• composition of generating functions of one variable
• composition of generating functions of two variables
• composition of generating functions of many variables
• mutual generating function
• mutual generating function of one variable
• mutual generating function of two variables
• mutual generating function of many variables
• inverse generating function
• inverse generating function of one variable
• inverse generating function of two variables variables
• inverse generating function of many variables
• pyramid
• tensor

Let the generating function $G(x)=\sum_{n>0} g(n)x^n$ be given and its composite. $$ G^{\Delta}(n,k)=[x^n]G(x)^k. $$ $G^{\Delta}(0,0)=1$. Then for coefficients of the composition of generating functions $A(x)=F(G(x))$ will be the formula is valid Look at the examples

Find an explicit expression for the Fibonacci numbers. $$ A(x)=\frac{x}{1-x-x^2} $$ Let 's imagine the generating function as a composition $$ A(x)=F(G(x)). $$ where $F(x)=\frac{1}{1-x}$, $G(x)=x+x^2$, $f(n)=1$, $G^{\Delta}(n,k)={k\choose n-k}$ $$ a(n)=\sum_{k=0}^n {k\choose n-k} $$

Consider obtaining an explicit expression for the expansion coefficients $A(x)=\sinh{\left(\frac{x}{1-x}\right)}$. It is known that $$ \sinh(x)=\sum_{n>0} \frac{1-(-1)^n}{2}\frac{x^n}{n!} $$ It is also known that $$ \left(\frac{x}{1-x}\right)^k=\sum_{n\geqslant k} Tp0002(n-k,k)x^n. $$ where (See encyclopedia) $$ Tp0002(n,k)={n+k-1 \choose n}. $$ Then $$ a(n)=\sum_{k=0}^n {n-1 \choose n-k} \frac{1-(-1)^k}{2\,(k!)} $$

To find the decomposition of the composition $A(x)=e^{\sin(x)}$ it is necessary to know composite $\sin(x)$ (in the encyclopedia it is written under the number Tp0114) $$G^{\Delta}(n,k)={\frac{(1+(-1)^{n-k})}{2^{k}\,n!} \sum_{i=0}^{\lfloor {k\over{2}}\rfloor}{(2\,i-k)^{n}{k\choose{i}}\left(-1\right)^{-i+{k+n\over(2)}}}}$$ Then $$ a(n)=\sum_{k=0}^n {\frac{(1+(-1)^{n-k})}{2^k n! k!} \sum_{i=0}^{ \lfloor {k\over{2}}\rfloor}{(2\,i-k)^{n}{k\choose{i }}(-1)^{-i+{k+n\over{2}}}}} $$

Up0002(x)/(1-x\,Up0020(x)) Let the generating function be given $${1-2,x}\over{1-3\,x+x^2}$$ Let 's rewrite it in the form $$ \frac{x}{1-x}\frac{1}{1-\frac{x}{(1-x)^2}}+1 $$ Consider the composition of functions $A(x)=\frac{x}{(1-x)^2}$ and $\frac{1}{1-x}$ The composite of the function $A(x)$ has the expression $$ A^{\Delta}(n,k)={n+k-1\choose n-k} $$ Then $$ [x^n]\frac{1}{1-A(x)}=\sum_{k=0}^n {n+k-1\choose n-k} $$ $$ \frac{x}{1-x}\frac{1}{1-A(x)}=\frac{x}{1-x}\left(1+\frac{x}{(1-x)^2}x+\frac{x^2}{(1-x)^4}x^2+\ldots+\frac{x^n}{(1-x)^{2n}}x^n+\ldots\right)= $$ $$ =x\,\left(\frac{1}{1-x}+\frac{x}{(1-x)^3}x+\frac{x^2}{(1-x)^5}x^2+\ldots+\frac{x^n}{(1-x)^{2n+1}}x^n+\ldots\right)= $$ the series written in parentheses has odd degrees $2n+1$. From where the summation formula will have the form. $$\sum_{k=0}^{n}{n+k\choose{n-k}}$$ Taking into account shift by $x$ and plus 1 (see the original formula, we get the desired expression). $$[x^n]{1-2,x\over{1-3\,x+x^2}}=\sum_{k=0}^{n}{n+k-1\choose{2k}}$$

Let be given a generating function for harmonic numbers $$H(x)=\frac{1}{1-x}\log\left(\frac{1}{1-x}\right)$$ Consider the composition functions $H(1-e^x)$. After substitution and transformations we get $$ H(1-e^x)=-x\,e^x. $$ Knowing that the composite of the function $A(x)=e^x-1$ has the expression $$ A^{\Delta}(n,k)= {n\brace k} \frac{k!}{n!} $$ then $$ [x^n](1-e^x)^k=(-1)^k\ {n \brace k} \frac{k!}{n!} $$ From where using the formula of the composition will get $$ [x^n]H(1-e^x)=\sum_{k=0}^n (-1)^k\ {n \brace k} \frac{k!}{n!}H_k, $$ where $H_k$ - harmonic numbers. On the other hand $$ [x^n]e^{-x}(-x)=\frac{(-1)^n}{(n-1)!}. $$ Now equate the left and the right part $$ \sum_{k=0}^n (-1)^k\ {n \brace k} \frac{k!}{n!}H_k=\frac{(-1)^n}{(n-1)!} $$ Let 's make transformations and we get the identity $$ \sum_{k=0}^n (-1)^{n-k}\,k!\, {n \brace k} \,H_k=n $$

${\bf Example}$ Let $$ G(x)=\frac{\log(1+x)}{x} $$ and $$$ G(n,k)=\frac{k!}{(n+k)!}{n+k \brack k}(-1)^n. $$ Let 's write down the equation $$ A(x)=G\left(\frac{x}{A(x)}\right)=\frac{\log(1+\frac{x}{A(x)}}{\frac{x}{A(x)}} $$ From where, we get the generating function for the Bernoulli numbers $$ A(x)=\frac{x}{e^x-1} $$ Then $$ A(n,k)=k\,\sum_{j=1}^{n}\frac{1}{j} {\left(-1\right)^(n+j-1)}\frac{j!}{(n+j)!} {n+j \brack j}{n-k+j-1\choose j-1} {2,n-k \choose n-j} $$ $A(0,k)=1$

Let the generating function $G(x)=\sum_{n\geqslant 0} g(n)x^n$be given and the coefficients of expansion of its degree $$ G(n,k)=[x^n]G(x)^k $$ Then the generating function for the diagonal $G(n,n)$ will have an expression $$ \frac{x\,A'(x)}{A(x)}, $$ where $A(x)$ is the solution of the functional equation $$ A(x)=G(x\,A(x)). $$ If $A(n,k)$ is known, then the diagonal $G(n,n)$ will have the following expression Example. Let the generating function $$ G(x)=\frac{1+x-x^2}{1-x} $$ be given and the expression for the coefficients of the degree is given $$ [x^n]G(x)^k=G(n,k)=\sum_{j=0}^{\left \lfloor {n \over 2} \right \rfloor}{k \choose j}\,{n-j-1\choose n-2\,j} $$ It is necessary to find the generating function for the diagonal $G(n,n)$. To do this , you need to solve the equation $$ A(x)=G(x\,A(x)=\frac{1+x\,A(x)-x^2\,A(x)^2}{1-A(x)} $$ $$A(x)^2\,\left(x^2+x\right)+A(x),\left(-x-1\right)+1=0$$ $$A(x)=1+x-\sqrt {1-2x-3x^2}\over{2\,x+2\,x^2}$$ We find the logarithmic derivative $$B(x)=\frac{(xA(x))'}{A(x)}\frac{1}{2} ({1\over 1+x}+{1\over\sqrt{1-2x-3x^2}}) $$ from where $$ [x^n]B(x)=G(n,n)=\sum_{j=0}^{\left \lfloor {n \over 2 } \right \rfloor}{n \choose j}\,{n-j-1 \choose n-2\,j} $$

The consequence of the statements of the paragraph 8.9 is that for the composition of the functions of $F(x)=\log(A(x))$, the expansion coefficients will have the expression Let's look at the examples
Example 1 $$ \log\left(\frac{1}{1-x}\right) $$ $$ [x^n]\left(\frac{1}{1-x}\right)^k={n+k-1\choose n} $$ $$\sum_{j=1}^{n}\frac{(-1)^{j-1}}{j}{n+j-1\choose n}{n \choose j}=\frac{1}{n}$$ Example 2 $$\log(e^x)=x$$ $$\sum_{j=1}^{n} (-1)^{j-1}\,{n\choose j}j^{n-1}=\left\{\begin{array}{ll} 1,& n=1,\ 0,& n>0. \end{array}\right.$$

Table of formulas for determining coefficients of degrees of series composition

N	Composition $G(x,y)^k$	The formula for $g(n,m,k)$
1	$G(x,y)^k=H(A(x),y)^k$	$g(n,m,k)=\sum\limits_{q=0}^n A^{\Delta}(n,q)h(q,m,k)$
2	$G(x,y)^k=H(x,B(y))^k$	$g(n,m,k)=\sum\limits_{r=0}^m B^{\Delta}(m,r)h(n,r,k)$
3	$G(x,y)^k=H(A(x,y))^k$	$g(n,m,k)=\sum\limits_{q=0}^{n+m} A^{\Delta}(n,m,q)h(q,k)$
4	$G(x,y)^k=H(A(x,y),y)^k$	$g(n,m,k)=\sum\limits_{q=0}^{n+m}\sum\limits_{r=0}^{m}{A^{\Delta}\left(n ,m-r,q\right)h(q,r,k)}$
5	$G(x,y)^k=H(x,B(x,y))^k$	$g(n,m,k)=\sum\limits_{q=0}^{n}{\sum\limits_{r=0}^{n+m-q}{B^{\Delta}\left(n-q ,m,r\right)}\,h(q,r,k)}$
7	$G(x)^k=H(x,B(x))^k$	$g(n,k)=\\sum\\limits_{q=0}^{n}{\\sum\\limits_{r=0}^{n-q}{B^{\\Delta}(n-q,r)\, h(q,r,k)}}$

$$G(x,y)={x,y-1\over{x,y+x^2+x-1}}$$ it can be represented as $$ G(x,y)=\frac{1}{1-\frac{x+x^2}{1-x\,y}}=H(A(x,y)), $$ where $H(x)=\frac{1}{1-x}$, $A(x,y)=\frac{x+x^2}{1-x\,y}$ In the knowledge base we do request for a generating function $U(x,y)=\frac{1+x}{1-y}$. We will get the pyramid found at number 17. Then the formula $x\frac{1+x}{1-y}$ will be true for the composites of the function $$ T_{17}(n-k,m,k)={k\choose{n-k}}\,{m+k-1\choose{m}} $$ Now let's do let's do the product of the variable $y$ by $x$ we get $$ A^{\Delta}(n,m,k)=T_{17}(n-m-k,m,k)={k\choose{n-k}}\,{m+k-1\choose{m}} $$ From where, based on the formula of the composition of two variables (see Table 1, line 3, $k=1$) $$ g(n,m)=\sum_{k=0}^{n+m} A^{\Delta}(n,m,k)=\sum_{k=0}^{n-m}{k\choose{n-m-k}}\,{m+k-1\choose{m}}. $$ Thus, we have obtained an explicit formula for the triangle sequence A055830, describing the class of paths on the lattice [12].

Let be given a generating function of the form: $$G(x,y)={x-\sqrt{x^2-2\,x^2\,y+2\,\sqrt{1-4\,x}\,x^2\,y}}\over{2\,x^2\,y}$$ Let 's find an explicit formula for the mutual generating function: $$ A(x,y)=\frac{1}{G(x,y)} $$ Decompose the function $G(x,y)$ nto a Taylor series at the point $x=0$ and $y=0$ $$ \begin{array}{l}\ 1\ +\left(1+y+\cdots \right)\,x\ +\left(2+2\,y+2\,y^2+\cdots \right)\,x^2\ +\left(5+5\,y+6\,y^2+5\,y^3+\cdots \right)\,x^3\ +\left(14+14\,y+18\,y^2+20\,y^3+14\,y^4+\cdots \right)\,x^4+\cdots \end{array}$$ Note that this function describes a triangle and $G(0,0)=1$. Then we write the function $H\left(x,\frac{y}{x}\right)$. Looking for this function in the knowledge base, we get the function and tensor number 69 (see paragraph 10) Then the expression will be true for the original function: $$ G(x,y)=\sum_{n}\sum_{m} T_{69}(n-m,m,k)x^n\,y^m. $$ To obtain an expression of the coefficients of the mutual generating function, you can use the expression[10]: $$T_r(n,m,k)=\sum_{j=0}^{m+n}{T(0 , 0 , 1)^{-j-k}(-1)^{j}{j+k-1\choose{j}}{\it T}\left(n , m , j\right) {k+m+n\choose{m+n-j}}}$$ Where $T(n,m,k)$ - is the tensor for the original generating function, provided that $T(0,0,0)=1$ and $T(n,m,0)=0$. Write down the tensor for the original function $$ T(n,m,k)=T_{69}(n-m,m,k)={\frac{k}{n+k}{2\,m+k-1\choose{m}}{2\,n+k-1\choose{n-m}}}. $$ Then the tensor of the mutual generating function will have an explicit expression $$ T_r(n,m,k)=\sum_{j=0}^{m+n}(-1)^{j} {j+k-1\choose{j}} {\frac{j}{n+j}} {2\,m+j-1\choose{m}} {2\,n+j-1\choose{n-m}} {k+m+n\choose{m+n-j}}. $$

Let $$ G(x,y)^k=\sum_{n\geqslant 0}\sum_{m\geqslant 0} T(n,m,k)x^ny^m $$ $G(0,0)\neq 0$.Then the coefficients of the decomposition of the function $$ \frac{1}{G(x,y)^{\alpha}} $$ will have the following expressions and Let's look at the example $$ \frac{x+y}{\log(1+x+y)} $$ $$ (\log(1+x)^k=\sum_{n\geqslant k} (-1)^{n-k}{n \brack k} \frac{k!}{n!}\,x^n $$ $$ \left(\frac{\log(1+x)}{x}\right)^k=\sum_{n\geqslant 0} (-1)^n\, {n+k \brack k} \frac{k!}{(n+k)!}x^n $$ Let's find the composition $$ A(B(x,y))=\frac{\log(1+x+y)}{x+y} $$ where $A(x)=\log(1+x)$, $B(x,y)=x+y$. Let's find the function $B(x,y)$ in the encyclopedia.This function is written by number 0001. Where from $$ B(x,y)^k= \sum_{n\geqslant 0} \sum_{m \geqslant 0} \delta_{k, n+m}\, {n+m \choose m }\,x^n\,y^m $$ Then we use the formula compositions (see table). We get $$ \sum_{j=0}^{n+m} B^{\Delta}(n,m,k)A^{\Delta}(j,k)= $$ $$ \sum_{j=0}^{n+m} \delta_{j, n+m} \,{n+m \choose m }(-1)^j\,{j+k \brack k}\frac{k!}{(j+k)!}= $$ $$ {n+m \choose m}(-1)^{n+m} {n+m+k \brack k} \frac{k!}{(n+m+k)!} $$ $$ \sum_{j=0}^{n+m} {\left(-1\right)^{j}\,}\ { j+\alpha-1 \choose j} {n+m \choose m} (-1)^{n+m} {n+m+j \brack j} \frac{j!}{(n+m+j)!}\, {n+m+\alpha \choose n+m-j} $$

Let the generating function $$ F(x,y)=\frac{x}{1-x-x^2(1+y)} $$ be given. Find the coefficients of the expansion of the inverse function satisfying the equation $$ G(A(x,y),y)=x . $$ To do this , we write $$ G_x(x,y)=\frac{G_x(x,y)}{x}. $$ Then the equation will take the form: $$ A(x,y)=\frac{x}{G_x(A(x,y),y)} $$ Now let's perform the substitution $A(x,y)=xA_x(x,y)$ $$ A_x(x,y)=\frac{x}{G_x(A_x(x,y),y)} $$ Then to find the inverse of the functions, you can use the formula. However, in this case the mutual mutual function is easy to find $$ G_{r}(x,y)=1-x-x^2(1+y)=(1-x(1+x+xy). $$ Now we find the pyramid $T(n,m,k)$ of the function $F(x,y)=(1+x+xy)$ at number 37 (see the fragment below). $$ T_r(n,m,k)=\sum_{i=0}^{n+m} T_{37}(n-i,m,i){k\choose i}(-1)^i=\sum_{i=0}^{n+m} {i\choose{n-i}}\,{n-i\choose{m}}{k\choose i}(-1)^i= $$ $$ \sum_{i=0}^{n} {i\choose{n-i}}\,{n-i\choose{m}}{k\choose i}(-1)^i $$ Then the coefficients for $A_x(x,y)^k$ $$ T_{A_x}(n,m,k)=\frac{k}{n+k}T_r(n,m,n+k)=\frac{k}{n+k}\sum_{i=0}^{n} {i\choose{n-i}}\,{n-i\choose{m}}{k+n\choose i}(-1)^i $$ $${\sqrt{4\,x^2\,y+5\,x^2+2\,x+1}-x-1}\over{2\,x^2\,y+2\,x^2}$$

Let the expression be given $$\sum_{k=0}^{m}{n+m-k+2\choose{k}}{n+m-k\choose{n}}$$ We will perform a number of transformations in order to obtain one of the compositional formulas presented in [3]. $$\sum_{k=0}^{m}{n+m-k+2\choose{ k}}{m-k+n\choose{n}}$$ changing the summation order from $k$ to $m-k$ $$\sum_{k=0}^{m}{n+k+2\choose{ m-k}}{k+n\choose{n}}$$ now the index $k$ does not start from zero, but from $n$ $$\sum_{k=n}^{n+m}{k+2\choose{ n+m-k}}{k\choose{n}}$$ now the index k $k$ starts from zero. $$\sum_{k=0}^{n+m}{k+2\choose{ n+m-k}}\,{k\choose{k-n}}$$ we introduce a new variable and introduce summation by this variable, using the kronecker symbol $$ j=n+m-k $$ we get $$\sum_{k=0}^{n+m}{\sum_{j=0}^{m}{\delta_{j, n+m-k} \,{k+2\choose{ j}}\,{-j+n+m\choose{m-j}}}}$$ now we compare it with the compositional formula $$g(n,m)=\sum\limits_{k=0}^{n+m}\sum\limits_{j=0}^{m}{A^{\Delta}\left(n ,m-j,k\right)h(k,j)}$$ Note that for our case $$ A^{\Delta}\left(n ,m-j,k\right)={\delta_{j, n+m-k}{-j+n+m\choose{m-j}}} $$ Where from $$ A^{\Delta}(n ,m,k)={\delta_{k, n+m}{n+m\choose{m}}} $$ From where, using the encyclopedia, we will find the pyramid at number 1, its generating function is $(x+y)$. Now let's find the generating function for $$ h(n,m)={n+2\choose{ m}} $$ We make a request to the knowledge base: $T=binomial(n+2,m)$ We get the pyramid number 42 (see the fragment below). From where we get the desired generating function. $$ A(x,y)=UU0042(x+y,y)= {\left(y+1\right)^2 \over 1-\left(y+1\right)\,\left(y+x\right)} $$

1. Let be given a functional equation of the form: $$ A(x,y)=G(x\,A(x,y),y) $$ And the expression of the coefficients of the k-degree of the generating function $G(x,y)$. $$ T_G(n,m,k)=[x^n\,y^m]G(x,y)^k. $$ is known. Then the coefficients of the logarithmic partial derivative of the generating function $A(x,y)$ in $x$ , will be expressed in terms of the coefficients $T_G(n,m,k)$ $$ T_{lx}(n,m)=[x^n\,y^m]\frac{1}{A(x,y)}\frac{\partial A(x,y)}{\partial x} $$ $$ T_{lx}(n,m)=T_G(n,m,n) $$ 2.Let a functional equation of the form be given: $$ A(x,y)=G(x,y\,A(x,y)) $$. And the expression of the coefficients of the k-degree of the generating function is known $G(x,y)$. $$ T_G(n,m,k)=[x^n\,y^m]G(x,y)^k. $$ then the coefficients of the logarithmic partial derivative of the generating function $A(x,y)$ in $y$ , will be expressed in terms of the coefficients $T_G(n,m,k)$ $$ T_{ly}(n,m)=[x^n\,y^m]\frac{1}{A(x,y)}\frac{\partial A(x,y)}{\partial y} $$ $$ T_{ly}(n,m)=T_G(n,m,m) $$ Let's find the coefficients of the logarithmic derivative of the generating function $A_{139}(x,y)$ of the corresponding frame in the knowledge base shown in the fragment below: Next, follow the link to the frame 59 and copy the explicit formula for the tensor of the generating function UU0059(x,y). $$ T_{59}(n,m,k)=\frac{k {\binom{2 k}{n}} {\binom{k + m}{m}}}{k + m} $$ For the desired logarithmic derivative $$ G(x,y)=\frac{1}{A(x,y)}\frac{\partial A(x,y)}{\partial x}= 1-2\,x^2-{2\,x^2\over{U(x\,y)}}+{x^2\,(4-4\,y)\over{2{\it U}\left(x , y\right)\,\sqrt{1-4\,x-\left(2-4\,x\right)\,y+y^2}}}, $$ where $$ U(x,y)=1-2\,x-y-\sqrt{1-4\,x-\left(2-4\,x\right)\,y+y^2}. $$ the coefficients will have an explicit expression $$ g(n,m)=T_{59}(n,m,n)={n\,{2\,n\choose{n}}\,{m+n\choose{m}}\over{m+n}}=\binom{2n}{n}\binom{n+m-1}{m}. $$ 3. Consider the case when the left tensor for this generating function is unknown, but there is a tensor of this function. Then you can use the following formulas. The formula for the partial logarithmic derivative of $x$ variable is as follows: $$ T_{logx}(n,m)=\left\{\begin{array}{ll}\ T(0,0,1)^n & n=0\;, m=0,\ n\,\sum_{j=1}^{n+m}{(-1)^{j-1}\,T(n,m,j)\,{n+m\choose{j}} \over{T^{j}(0 , 0 , 1)j}}, & n>0\;or\; m>0.\ \end{array} \right. $$ The formula for the partial logarithmic derivative of $x$ variable is as follows: $$ T_{logy}(n,m)=\left\{\begin{array}{ll}\ T(0,0,1)^m & n=0\;, m=0,\ m\, \sum_{j=1}^{n+m}{(-1)^{j-1}\, T(n , m , j){n+m\choose{j}}} \over{T^{j}(0 , 0 , 1)\,j}, & n>0\;or\; m>0.\ \end{array} \right. $$ Comparing the above formulas for partial logarithmic derivatives, you can see that they differ by multipliers $m$ and $n$. The terms $T_{logx}(n,m)$ and $T_{logy}(n,m)$ describe the decomposition of the logarithmic derivative if the terms $T_{logx}(n,m)$ by $n$, and $T_{logy}(n,m)$ by $m$, we obtain the terms of the decomposition of the composition of generating functions $$ log(U(x,y)), $$ where $U(x,y)$- is the generating function having the tensor $T(n,m,k)$.Let 's write down an explicit formula for the members of the composition $$ Lo(n,m)=\sum_{j=1}^{n+m}{(-1)^{j-1} T(n , m , j){n+m\choose{j}} \over{T^{j}(0 , 0 , 1)j}} $$ Consider an example. Let the generating function number 130 be given. Then the coefficients of the decomposition of the composition of functions $$ \log(UU_{130}(x,y)=\log({1\over{-4\,y+(1-x+y)^2}}) $$ $$ Lo(n,m)=\sum_{k=1}^{n+m} (-1)^{k-1} { {m+k\choose{k-1}} {n+k-1\choose{k-1}} {n+m+2\,k-1\choose{k-1}} {2\,(n+m)+2\,k\choose{2\,m+1}} \over 2\,k\,{2\,k-2\choose{k-1}}\,{2\,m+2\,k\choose{2\,k-2}} } {n+m\choose{k}} $$

Consider the generating function for the Fin triangle $$ G(x,y)={2\over{1+\sqrt{1-4\,x}+2\,x-2\,x\,y}} $$ $$G(x,y)=\frac{A(x)}{1-y\,A(x)}.$$ $$A(x)=\frac{2}{1+\sqrt{1-4\,x}+2\,x}$$ $$ G(x,y)=A(x)\,y+A(x)^2\,y^2+\ldots+A(x)^n\,y^n+\ldots $$ $A(x)$ - s the generating function of the Fin sequence (see. $Up0157(x)$). Then $$ G(x,y)=\sum_{n\geqslant k} Tp0157(n,k)\,x^n\,y^k $$

Generating function for Euler numbers of the first genus $$E(x,y)={y-1\over{y-e^{x(y-1)}}}$$ Let 's imagine it in the form $$E(x,y)={1\over{1-\frac{e^{x,(y-1)}-1}{y-1}}}$$ Consider the function $$ A(x,y)=\frac{e^{x(y-1)}-1}{y-1} $$ Then it can be represented by a composition of functions $$ A(x,y)=x\,B(C(x,y)), $$ where $B(x)=\frac{e^x-1}{x}$ and $C(x,y)=x\,(y-1)$. Then $B(x)=Up0102(x)/x$ $$ B(x)^k=(\frac{e^x-1}{x})^k=\sum_{n\geqslant 0} Tp0102(n+k,k)x^n. $$ where $$Tp0102(n,k)=\frac{k!}{n!}\,{n \brace k}$$ Using the encyclopedia we will find $C(x)^k$ $$ C(x)^k=\sum_{n\geqslant k}\,Tuu0002(n,m,k)\,(-1)^{m+k}\,x^n\,y^m $$ $$ Tuu0002(n,m,k)=\delta_{k, n} \,{n\choose m} $$ We find the coefficients $B(C(x))^k$ $$ \sum_{j=0}^{n+m} Tuu0002(n,m,j)(-1)^{m+j}Tp0102(j+k,k)= $$ $$ \sum_{j=0}^{n+m}\delta_{j, n} \,{n\choose m}(-1)^{m+j}\frac{k!}{(j+k)!}\,{j+k \brace k}= $$ $$ {n \choose m }(-1)^{n+m}\frac{k!}{(n+k)!}\,{n+k \brace k}. $$ We find the coefficients $x\,B(C(x))^k$ $$ {n-k\choose m }(-1)^{n+m-k}\frac{k!}{n!}\,{n \brace k}= $$ From where we find an explicit expression for the decomposition coefficients of the desired composition $$ \sum_{k=0}^{n} {n-k\choose m}(-1)^{n+m-k}\frac{k!}{n!}\,{n \brace k} $$ Then Euler numbers of the first kind will have the following explicit expression $$ {\left\langle \matrix{n\cr m} \right\rangle}=\sum_{k=0}^{n-m} (-1)^{n+m-k}\,k!\,{n-k \choose m} {n \brace k} $$

Consider the use of an encyclopedia to obtain a generating function for Euler numbers of the second genus. To do this, we prove the following theorem.
Theorem 2 The exponential generating function for second-order Euler numbers has the form: where $W(x)$ -is the Lambert function, $x>0$.
Proof We write down the exponential generating function in the form of a double series, while taking the offset by $(n-1)$ $$ E(x,t) = \sum_{n>0}\sum_{m>0} {\left\langle \left\langle \matrix{n-1\cr m} \right\rangle \right\rangle} \frac{x^n}{n!}t^m $$ Consider the formula obtained in [15] It can be represented as the product of two rows $$ E(x,t)=\sum_{n>0}P_n(t)\frac{x^n}{n!}\,(1-t)^{2n-1} $$ where $$ (1-t)^{2\,n-1}=\sum_{n\geqslant 0}(-1)^m\,{2\,n-1 \choose m}t^m $$ and $$ P_n(t)=\sum_{m\geqslant 0}{n+m-1 \brace m}t^m $$ Let 's write the function $$ u(x,t)=\sum_{n>0}\sum_{m\geqslant 0}{n+m-1 \brace m}\frac{x^n}{n!}t^m $$ We show that this function is the inverse function for the function $$ y(x,t)=(x-t(e^x-1)). $$ let 's write a mutual function for $y(x,t)$ $$ \frac{x}{y(x,t)}=\frac{1}{1-t\frac{e^x-1}{x}} $$ and find her degree. To do this , we write $$ (t(\frac{e^x-1}{x}))^k=\sum_{n>0} \sum_{m>0}T(n,m,k)x^nt^n $$ It is known that for the function $(e^x-1)$ is available identity $$ (e^x-1)^k=\sum\limits_{n\geqslant 0} k!{n \brace k}\frac{x^n}{n!}, $$ we get $$ T(n,m,k)=\delta(m,k){n+k \brace k}\frac{k!}{(n+k)!}, $$ where $\delta(m,k)$ is the Kronecker symbol. For the function $\frac{1}{(1-x)}$ the identity is known $$ \frac{1}{(1-x)^k}=\sum_{n\geqslant 0} {n+k-1 \choose n}x^n. $$ Then for the degree of composition of two generating functions $$ ({1 \over 1-t\frac{e^x-1}{x}})^k =\sum\limits_{n\geqslant 0} \sum\limits_{m\geqslant 0} D(n,m,k)x^nt^m, $$ based on the compositional formula \cite{kru}, we can write $$ D(n,m,k)=\sum\limits_{i=0}^{n+m} T(n,m,i){i+k-1 \choose i}=\sum\limits_{k=0}^{n+m} \delta(m,i) {n+i \brace i}\frac{i!}{(n+i)!}{i+k-1 \choose i}= $$ $$ ={n+m \brace m}\frac{m!}{(n+m)!}{m+k-1 \choose m}. $$ Now let 's use the theorem Lagrange on the inverse function, according to which for a power series $u(x,t)$ satisfying the functional equation $$ u=x\,F(g,t) $$ where $F(x,t)$ - s a power series for which $F(0,0)\neq 0$ holds the identity. $$ [x^n]u(x,t)=\frac{k}{n}[x^{n-k}]F(x,t)^n. $$ Let 's write down the functional equation for our case $F(x,t)=\frac{1}{1-t\frac{e^x-1}{x}}$ $$ u=\frac{x}{1-t\frac{e^{u}-1}{u}}. $$ The solution to this equation will be $$ u(x,t)^k=\sum\limits_{n>0} \sum\limits_{m\geqslant 0} \frac{k}{n}D(n-k,m,n)x^n\,t^m $$ where $D(n,m,k)$ where are the coefficients of the series $F(x,t)^k$. Then $$ D(n-k,m,n)={n+m-k \brace m}\frac{m!}{(n+m-k)!}{m+n-1 \choose m} $$ For $k=1$ we get $$ u(x,t)=\sum\limits_{n>0} \sum\limits_{m\geqslant 0} \frac{1}{n}{n+m-1 \brace m}\frac{m!}{(n+m-1)!}{m+n-1 \choose m}x^n\,t^m $$ or, after simplifying the binomial coefficient and factorials, we get $$ u(x,t)=\sum\limits_{n>0} \sum\limits_{m\geqslant 0} \frac{1}{n!}{n+m-1 \choose m}x^n\,t^m $$ Thus, the function $u(x,t)$ is the inverse of the function $y(x,t)=(x-t(e^x-1))$. Now let's use the Lambert function and its properties. The Lambert function is given by the equation $$ x=W(x)e^{W(x)} $$ It is known that solutions of the equation of the form $$ g(x)=f(x)e^{f(x)} $$ the solution will be $$ f(x)=W(g(x)) $$ We use this property for our case $$ y=x(y,t)-t\,e^{x(t,y)}+t $$ We will make a replacement $$ z(y,t)=x(y,t)+t-y $$ Then the equation will have the form $$ z(y,t)=t\,e^{Z(y,t)-t+y} $$ Let 's rewrite it in the form $$ z(y,t)e^{-Z(y,t)}=t\,e^{y-t} $$ Then the solution for $z(y,t)$ will have the form $$ z(y,t)=-W(-t\,e^{y-t}) $$ Where from $$ x(y,t)=y-t-W(-t\,e^{y-t}) $$ Now consider the product $$ E(x,t)=\sum_{n>0}P_n(t)\frac{x^n}{n!}\,(1-t)^{2n-1}=\frac{1}{1-t}\sum_{n>0}P_n(t)\frac{(x(1-t)^2)^n}{n!} $$ Where from $$ E(x,t)=\frac{E_2(x(1-t)^2,t)}{(1-t)}=\frac{x(1-t)^2-t-W(-t\,e^{x(1-t)^2-t})}{1-t}. $$ Differentiating by $x$ the expression for $E(x,t)$ taking into account the properties of the derivative of the Lambert function $W(x)$ we obtain the desired generating function $$ Eu(x,t)=\sum\limits_{n>0} \sum_{m>0} {\left\langle \left\langle \matrix{n\cr m} \right\rangle \right\rangle} \frac{x^n}{n!}t^m={1-t\over{W(-t\,e^{(1-t)^2\,x-t})+1}} $$

Consider the composition of generating functions $F(x)$ and $G(x,y,\ldots,z)$, $G(0,0,\ldots,0)=0$ Then $A(x,y,\ldots,z)=F(G(x,y,\ldots,z))$ $$ \sum_{k=0}^{n+m+\ldots, i} G^{\Delta}(n,m,\ldots,i,k)f(k) $$ $$ [x^n\,y^m\,\ldots\,z^i](x+y+\ldots+z)^k={n+m+\ldots+i\choose n,m,\ldots,i}\delta(n+m+\ldots+i,k) $$ $$ [x^n\,y^m\,\ldots\,z^i]F(x+y+\ldots+z)=\sum_{k=0}^{n+m+\ldots+i}{n+m+\ldots+i\choose n,m,\ldots,i}\delta(n+m+\ldots+i,k)f(k)= $$ $$ {n+m+\ldots+i\choose n,m,\ldots,i}f(n+m+\ldots+i) $$ Example $$ e^{\frac{x}{1-x-y-z}} $$ $$ [x^n\,y^m\,z^i]\left(\frac{1}{1-x-y-z}\right)^k= $$ $$ \sum_{j=0}^{n+m+i} {n+m+i\choose n,m,i}\delta(n+m+i,j){j+k-1\choose j}= $$ $$ {n+m+i\choose n,m,i}{n+m+i+k-1\choose n+m+i} $$ Where from $$ [x^n\,y^m\,z^i]\left(\frac{x}{1-x-y-z}\right)^k={n-k+m+i\choose n-k,m,i}{n+m+i-1\choose n-k+m+i} $$ $$ [x^n\,y^m\,z^i]e^{\left(\frac{x}{1-x-y-z}\right)}=\sum_{k=0}^{n+m+i}\frac{1}{k!}{n-k+m+i\choose n-k,m,i}{n+m+i-1\choose n-k+m+i}= $$ $$ \sum_{k=0}^{n}\frac{1}{k!}{n-k+m+i\choose n-k,m,i}{n+m+i-1\choose n-k+m+i} $$

Let the function $G(x,y,\ldots,z)$, $G(0,0,\ldots,0)\neq 0$ and be given and the coefficients of the k-th degree are given $$ G(x,y,\ldots,z)^k=\sum_{n\geqslant 0}\sum_{m\geqslant 0} T(n,m,\ldots,i,k)x^n\,y^m. $$ Then for the coefficients of the expansion $G(x,y,\ldots,z)^{-\alpha}$ will have the expressions and

Consider the method of obtaining pyramids. o do this, we write down the set of generating functions of one variable and the corresponding set of explicit expressions of coefficients $k$-degrees. . Let there be a set of $G=\{g_i(x)\}$ generating functions, the coefficients of $k$-degree are described by binomial coefficients of $\{T_i(n,k)\}$. $$ \begin{array}{|c|c|}\hline \hbox{Generating function }G(x) &\hbox{Coefficients of degree } g_x(n,k)\\ \hline (1+x) & \binom{k}{n}\\ \hline \frac{1}{1-x} & \binom{n+k-1}{n}\\ \hline \frac{1-\sqrt{1-4x}}{2x} & \frac{k}{n+k}{2n+k-1\choose n}\\ \hline \frac{1-2x-\sqrt{1-4x}}{2x^2} & \frac{k}{n+k}{2n+2k\choose n}\\ \hline \frac{2}{\sqrt{3x}} \sin({ {\arcsin( {3^ {3\over{2}} \sqrt{x} \over {2} } )} \over {3} }) & {k\,{3\,n+k-1 \choose n }}\over{2\,n+k}\\ \hline 1+\sqrt{1+4x}\over{2} & k(-1)^{n-1}{2n-k-1 \choose n-1}\over{n}\\ \hline 1-\frac{4}{3} \sin ^2({ {\arcsin( {3^ {3\over{2}} \sqrt{-x} \over {2} } )} \over {3} }) & k(-1)^{n-1}{3n-k-1 \choose n-1}\over{n} \\ \hline \sqrt{1+4x} & \cases{ \begin{array} \binom {\binom{k \over 2}n} 4^{n} & k-even \\ \frac{(-1)^{n-{k+1 \over 2}} \binom {n} {k+1\over 2} \binom {2n} {n}} {\binom {2n}{k+1}}& k-odd, 2n>(k+1),\\ \frac{\binom{k+1}{2n}\binom {2n}{n}} {\binom{k+1\over 2} {n}} & k-odd,2n \le(k+1), \\ \end{array} }\\ \hline \sqrt{1+4x^2}+2x & \cases{ \begin{array}\\ \frac{k4^n}{n+k} \binom {n+k \over 2}{n},& n+k-even\\ \frac{k(-1)^{n-k-1\over2}\binom {n}{n+k+1 \over 2} \binom{2n}{n}}{(n+k) \binom{2n}{n+k+1}} &k-odd,n>(k+1),\\ \frac {k\binom{n+k+1}{2n} \binom{2n}{n}}{(n+k) \binom{n+k+1 \over 2}{n}}& n+k-odd,n\le(k+1),\\ \end{array} }\\ \hline {1\over \sqrt{1-4x}} & \cases{ \begin{array}\\ 4^n \binom {\frac{k}{2} + n-1}{n} & k-even, \\ \frac {\binom {2n+k-1}{\frac{k-1}{2}+n} \binom{\frac{k-1}{2}+n}{n}} {\binom{k-1}{k-1 \over 2}} ,& k-odd,\\ \end{array}} \\\hline {1\over 12x} + \frac{\sin (\frac{\arcsin(-1+216\,x^2)}{3})} {6x} & \cases{ \begin{array}\\ \frac{k4^n \binom {\frac{3n+k}{2}-1}{n}}{n+k}, & k - even,\\ \frac{k \binom{\frac{3n+k-1}{2}}{n} \binom{3n+k-1}{\frac{3n+k-1}{2}}} {(n+k) \binom{n+k-1}{\frac{n+k-1}{2}}}& k -odd \\\end{array} }\\ \hline \sqrt{\frac{1-\sqrt{1-16x}}{8x}} & \cases{ \begin{array}\\ \frac{k4^n \binom{\frac{4n+k}{2}-1}{n}}{2n+k} & k - even,\\ \frac{k \binom{\frac{4n+k-1}{2}}{n} \binom {4n+k-1} {\frac {4n+k-1} {2} }} {(2n+k)\binom{2n+k-1}{\frac{2n+k-1}{2}}} ,& k- odd.\\ \\ \end{array}} \\ \hline \sqrt{\frac{\sin (\frac{\arcsin(\sqrt {27x})}{3})} {\sqrt {3x}}} & \cases{ \begin{array}\\ \frac{k4^n \binom{\frac{6n+k}{2}-1}{n}}{4n+k} & k - even,\\ \frac{k \binom{\frac{6n+k-1}{2}}{n} \binom {6n+k-1} {\frac {6n+k-1} {2} }} {(4n+k)\binom{4n+k-1}{\frac{4n+k-1}{2}}} ,& k- odd.\\ \\ \end{array} }\\ \hline \end{array}$$ It should be noted that this table is significantly limited and is used as an example. Choose $g_1(x)\in G$ and $g_2(x)\in G$ and construct a new function of two variables. Two ways of obtaining pyramids of the generating function of two variables can be proposed:

1. Based on the product and composition of functions $g_1(x)\in G$ and $g_2(x)\in G$. Let $g_1(x)=(1+x)$, $g_2(x)=\frac{1}{1-x}$ and the coefficients of k-degree $g_1(x)$ have the form: $$ T_1(n,k)={k \choose n}, $$ for $g_2(x)$ $$ T_2(n,k)={n+k-1 \choose n}, $$ Let's write a new function of two variables $$ U(x,y)=g_1(x)\cdot g_2(y) $$ Then the pyramid of the function $U(x,y)$ will have the expression $$ T_U(n,m,k)=T_1(n,k)\cdot T_2(m,k) $$ Then $$ U(x,y)=\frac{1+x}{1-y} $$ From where, we get the following pyramid $$ T_U(n,m,k)={k\choose n}{m+k-1\choose m}. $$ This pyramid is listed in the encyclopedia as number 17. We will expand this method by composing generating functions for obtaining pyramids. Let's choose $g_1(x)\in G$ and $g_2(x)\in G$ and build a new function of two variables $$ U(x,y)=g_1(x\cdot g_2(y))\,g_2(y) $$ Then the pyramid of the function $U(x,y)$ will have the expression $$ T_U(n,m,k)=T_1(n,k)\cdot T_2(m,n+k) $$ In general, you can write a function of the form $$ U(x,y)=g_1(x\cdot g_2(y)^a)^b\,g_2(y)^c $$ where $a,b,c\in N$ Then the pyramid of the function $U(x,y)$ will have the expression $$ T_U(n,m,k)=p_1(n,bk)\cdot p_2(m,an+ck) $$ For example, for $g_1(x)=(1+x)$, $g_2(x)=\frac{1}{1-x}$ (see the previous example), we will set the values $a=2$, $b=2$, $c=1$. Then the function will take the form $$ U(x,y):=\frac{1}{1-y}\left(1+\frac{x}{(1-y)^2}\right)^2 $$ From where the pyramid will have the expression: $$ T_U(n,m,k)=T_1(n,2k)T_2(m,2n+k)=\binom{2k}{n}\binom{m+2n+k-1}{m} $$ This pyramid is recorded in the encyclopedia under the number 1423.

2. Consider a method for obtaining pyramids based on Lagrange's theorem. Let 's choose $g_1(x)\in G$ and $g_2(x)\in G$ and write a functional equation of the form: $$ U(x,y)=g_1(x\cdot g_2(y)^a\cdot U(x,y))^b\,g_2(y)^c $$ To do this, you need to find the coefficients of the degree $$ T(n,m,k)=[x^n\,y^m]\left(g_1(x\cdot g_2(y)^a)^b\,g_2(y)^c\right)^k=T_1(n,b\,k)T_2(m,a\,n+c\,k) $$ To do this, use the previous method. Where from $$ T_U(n,m,k)=\frac{k}{n+k}T_1(n,b(n+k))\cdot T_2(m,an+c(n+k)) $$ Let's take the initial data from the previous example. Then the functional equation will take the form: $$ U(x,y)=\left(1+\frac{x\,U(x,y)}{1-y)^2}\right)^2\frac{1}{1-y}. $$ As a result of the solution , we obtain the following generating function of two variables: $$ U(x,y)=\frac{(1-\sqrt{1-4x(1-y)^3}-2x) (1-y)^2}{2x^2} $$ Then $$ T_F(n,m,k)= \frac{k}{n+k} \binom{2n+2k}{n} \binom{3n+m+k-1}{m} $$ This pyramid is recorded in the encyclopedia under the number 1458.
Using this technique, you can get a large number of pyramids. Currently, based on the use of this technique, 1502 pyramids have been obtained.