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Benedict Irwin edited untitled.tex
about 8 years ago
Commit id: 60fe039446f979359f8e769c0beddaae6e147d63
deletions | additions
diff --git a/untitled.tex b/untitled.tex
index c004d04..1772e55 100644
--- a/untitled.tex
+++ b/untitled.tex
...
a_k = 2^k\\
a_{k-1} = -k2^{k-1} \\
a_{k-2} = k(k+3)2^{k-3} \\
a_{k-4} a_{k-3} = \frac{1}{3}(k(38+k(k+9))2^(k-4)) \\
a_{k-5} a_{k-4} = \frac{1}{3}(378+179k+18k^2+k^3)2^(k-7)) \\
a_{k-6} a_{k-5} = \frac{1}{15}(k(9864+k(3030+k(515+k(30+k))))2^(k-8)) \\
\end{equation}
However we must shift the series along to deal with the fact that $S_k(n)$ has $k$ terms, and we end up with \begin{equation}
a_k = 4\cdot2^{k-2}\\
a_{k-1} = -(k-1)2^{k-3} \\
a_{k-2} = (k-2)(k+1)2^{k-5} \\
a_{k-3} = \frac{1}{3}((k-3)(38+(k-3)((k-3)+9))2^(k-7))
\end{equation}
The we can see that the power of $2$ always seems to contain the next prime.