Funções Geradoras (Cheat Sheet)

Fórmulas

U(z) = \sum_{k=0}^{+\infty}u_kz^k

Sucessão	Polinómio $U(z)$	Geradora $G(z)$	Termo Geral $u_k$
$1,1,1,1,1,\dots$	$1+z+z^2+...+z^n$	$\displaystyle\frac{1}{1-z}$	$1$
$1,2,3,4,5,\dots$	$1+2z+3z^2+\dots+\allowbreak{(n+1)z^n}$	$\displaystyle\frac{1}{(1-z)^2}$	$k + 1$
$0,1,2,3,4,\dots$	$0+1z+2z^2+\dots+\allowbreak{n\cdot z^n}$	$\displaystyle\frac{z}{(1-z)^2}$	$k$
$1,0,1,0,1,\dots$	$1+z^2+z^4+\dots+z^{2n}$	$\displaystyle\frac{1}{1-z^2}$	$1$ de 2 em 2
$1,0,0,2,0,0,\allowbreak3,0,0,4,\dots$	$1+2z^3+3z^6+\dots$	$\displaystyle\frac {1}{(1-z^3)^2}$	$k$ de 3 em 3
$1,a,a^2,a^3,a^4,\dots$	$1+az+a^2z^2+a^3z^3+\dots$	$\displaystyle\frac{1}{1-az}$	$a^k$
$1,1,1,1,0,0,0,0,\dots$	$1+z+z^2+z^3$	$\displaystyle\frac{1-z^4}{1-z}$	$1$ até $k=3$
$0,0,1,1,1,1,\dots$	$z^2+z^3+z^4+\dots$	$\displaystyle\frac{z^2}{1-z}$	$1$ para $k > 1$
$0,1,1,1,0,0,\dots$	$z+z^2+z^3$	$\displaystyle\frac{z(1-z^3)}{1-z}$	$1$ para $0< k<4$

\frac 1 {(1-\lambda z^p)^m} = \sum_{k=0}^{+\infty} {k+m-1 \choose m-1}\lambda^kz^{pk}

Teorema do Sudoeste do Triângulo de Pascal

\text{Para todo o } n, m \in \N \text{ tem-se:} \sum_{k=0}^{n} {k+m \choose m} = {n+m+1 \choose m+1}

Soma dos primeiros termos de uma sucessão

S(z)=\frac{U(z)}{1-z}

Teorema 3 (Alguém que dê nome a isto)

\frac 1 {(1-\lambda z)^m} = \sum_{k=0}^{+\infty} {k+m-1 \choose m-1}\lambda^kz^k

$l$ = valor da soma
$n$ = número de incógnitas
$x$ = incógnitas

x_1+x_2+...+x_n = l \Leftrightarrow {l+n-1 \choose n-1}

x_1+x_2+...+x_n \leqslant l \Leftrightarrow {l+n \choose n}

\text{Para restrições onde } x_i \geqslant l_i \\ \text{Se } x_1 \geqslant 2 ,\text{ então } l_1 = 2\\ \text{Se } x_1 \geqslant 4 ,\text{ então } l_1 = 4

x_1+x_2+...+x_n = l \Leftrightarrow {l - (l_1 +l_2 +...+l_n)+ n - 1 \choose n - 1}

Para $h_0 = 1$ e $h_1 = 2$

h_n=h_{(n-1)} + 2h_{\color{green}(n-2)}+3^n

{\sum_{k=\color{green}2}^{+\infty}h_kz^k}=z{\sum_{k=\color{green}1}^{+\infty}h_{k-1}z^{k-1}}+ 2z^2{\sum_{k=\color{green}0}^{+\infty}h_{k-2}z^{k-2}}+ {\sum_{k=\color{green}2}^{+\infty}3^kz^k}

H_n - 1 - 2z= z(H_n - 1) + 2z^2H_n + \frac{1}{(1-3z)} - 3^0 - 3z \\ H_n(1-z-2z^2) = \frac{1}{(1-3z)} - 2z \\ H_n= \frac{1}{(1-3z)(1-z-2z^2)} - \frac{2z}{(1-z-2z^2)}

P.P.S It ain´t Much but it's Honest Work 👨‍🌾