Benedict Irwin edited untitled.tex  about 8 years ago

Commit id: 7a0a87892bf67888d0d7e6944c26edb93adeae6f

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a_{k-5} = \frac{1}{3}(378+179k+18k^2+k^3)2^(k-7)) \\  a_{k-6} = \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} \\