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| author | mzuenni <michi.zuendorf@gmail.com> | 2024-09-10 16:03:27 +0200 |
|---|---|---|
| committer | mzuenni <michi.zuendorf@gmail.com> | 2024-09-10 16:03:27 +0200 |
| commit | cf3bf7ef7a3c4899b31041b15bcc67f6607f5cf4 (patch) | |
| tree | 4a6e419a4873b31cf58a1bfa71230b31ec63a940 /content/math/math.tex | |
| parent | fb5d046c4e4ab2bec7b0642626e11cff5680f63f (diff) | |
fix pentagonal number theorem
Diffstat (limited to 'content/math/math.tex')
| -rw-r--r-- | content/math/math.tex | 2 |
1 files changed, 1 insertions, 1 deletions
diff --git a/content/math/math.tex b/content/math/math.tex index c07a41e..bad3bad 100644 --- a/content/math/math.tex +++ b/content/math/math.tex @@ -485,7 +485,7 @@ Die Anzahl der Partitionen von $n$ mit Elementen aus ${1,\dots,k}$. \begin{align*}
p_0(0)=1 \qquad p_k(n)&=0 \text{ für } k > n \text{ oder } n \leq 0 \text{ oder } k \leq 0\\
p_k(n)&= p_k(n-k) + p_{k-1}(n-1)\\[2pt]
- p(n)&=\sum_{k=1}^{n} p_k(n)=p_n(2n)=\sum\limits_{k\neq0}^\infty(-1)^{k+1}p\bigg(n - \frac{k(3k-1)}{2}\bigg)
+ p(n)=\sum_{k=1}^{n} p_k(n)&=p_n(2n)=\sum\limits_{k\neq0}^\infty(-1)^{k+1}\bigg[p\bigg(n - \frac{k(3k-1)}{2}\bigg) + p\bigg(n - \frac{k(3k+1)}{2}\bigg)\bigg]
\end{align*}
\begin{itemize}
\item in Formel $3$ kann abgebrochen werden wenn $\frac{k(3k-1)}{2} > n$.
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