# Random stuff 2026 week 5
Laplace transform esque table for generating functions?
$\def \g{{\cal G}}$
If we denote ${\cal G}[a_{n}] = \sum_{n=0}^{\infty}a_{n}t^{n}$ as the generating function transform of the sequence $(a_{n})_{n=0}^{\infty}$, and by convention $a_{-k}=0$ for $k\ge 0$. Then we have:
$\g [1] = \frac{1}{1-t}$
$\g [n] = \frac{t}{(1-t)^{2}}$
$\g [r^{n}] = \frac{1}{1-rt}$
$\g [n a_{n}] = t \frac{d}{dt} \g [a_{n}]$
$\g[\frac{1}{n+1}a_{n}] = \int \g [a_{n}] dt$
$\g[\frac{1}{n} a_{n}] = \int t \g [a_{n}] dt$, $a_{0}=0$.
$\g [s_{+}(a_{n})] = \frac{1}{t} (\g [a_{n}]-a_{0})$
$\g [s_{++}(a_{n})] = \frac{1}{t^{2}}(\g [a_{n}] - a_{0} - a_{1}t)$
$\g \left[ \cos\left( n \frac{\pi}{2} \right) \right] = \frac{1}{1+t^{2}}$, note $(\cos(n \frac{\pi}{2})) = (\overline{1, 0, -1, 0})$.
$\g \left[ \sin\left( n \frac{\pi}{2} \right) \right] = \frac{t}{1+t^{2}}$, note $(\sin ( n \frac{\pi}{2})) = (\overline{0,1,0,-1})$.
$t \g [a_{n}] = \g [a_{n-1}:a_{-1}=0]$
$\frac{1}{t} \g [a_{n}] = \g [a_{n+1}]$, provided $a_{0} = 0$.
$\g [r^{n} a_{n}] = \g[a_{n}]|_{t\to rt}$.
$\g [\delta(n-k)] = t^{k}$, where $\delta(x) = 0$ if $x \neq 0$ and $= 1$ if $x = 0$.
$\g [u(n-k)] = \frac{t^{k}}{1-t}$, where $u(x) = 0$ if $x < 0$, and $u(x) = 1$ if $x \ge 0$.
$\g[u(n-k) a_{n-k}] = t^{k} \g[a_{n}]$
$\g [(\overline{b_{0},b_{1},\ldots,b_{k-1}})] = \frac{b_{0}+ b_{1}t + \cdots b_{k-1}t^{k-1}}{1-t^{k}}$.
$\g[a_{n} \ast b_{n}] = \g[a_{n}]\g[b_{n}]$, where $(a_{n}\ast b_{n})$ is their convolution, where the $n$-th term is $\sum_{k=0}^{n} a_{k}b_{n-k}$
where $s_{+}(a_{n}) = (a_{n+1})$ and $s_{++}(a_{n}) = s_{+}(s_{+}(a_{n}))$
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Lagrange points?
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Happy numbers and their cycles
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"Why Fish Don't Exist"
"To care for the hidden and insignificant."