Benedict Irwin edited untitled.tex  about 8 years ago

Commit id: 8cc516fd9cbae5964c7cb9700e943012e5ebc208

deletions | additions      

       

a_{k-1} = -(k)2^{k-1} \\  a_{k-2} = (3k+k^2)2^{k-3} \\  a_{k-3} = -\frac{2^{k-4}}{3}(38k+9k^2+k^3) \\  a_{k-4} = \frac{2^{k-7}}{3}(378+179k+18k^2+k^3) \frac{2^{k-7}}{3}(378k+179k^2+18k^3+k^4)  \\ a_{k-5} = -\frac{2^{k-8}}{15}(9864k+3030k^2+515k^3+30k^4+k^5) \\  a_{k-6} = \frac{2\cdot2^{k-11}}{45}(125640k+90634k^2+12915k^3+1165k^4+45k^5+k^6)\\  a_{k-7} = \frac{2\cdot2^{k-12}}{315}(1684080k+2003652k^2+463204k^3+40005k^4+2275k^5+63k^6+k^7) \frac{2\cdot2^{k-12}}{315}(1684080k+2003652k^2+463204k^3+40005k^4+2275k^5+63k^6+k^7)\\  a_{k-8} = \frac{2^{k-15}}{315}(42089040k+50017932k^2+14438676k^3+1728769k^4+101640k^5+4018k^6+84k^7+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}\\