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\end{equation}  In general we may note that for $S_k(n)$ \begin{equation}  a_k = 2^k\\  a_{k-1} = -k2^{k-1} -(k)2^{k-1}  \\ a_{k-2} = k(k+3)2^{k-3} (3k+k^2)2^{k-3}  \\ a_{k-3} = \frac{2^{k-4}}{3}(38k+9k^2+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) \\  a_{k-5} = \frac{2^{k-8}}{15}(9864k+3030k^2+515k^3+30k^4+k^5) -\frac{2^{k-8}}{15}(9864k+3030k^2+515k^3+30k^4+k^5)  \\ a_{k-6} = \frac{2\cdot2^{n-11}}{45}(125640k+90634k^2+12915k^3+1165k^4+45k^5+k^6)  \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} (3(k-2)+(k-2)^2)2^{k-5}  \\ a_{k-3} = \frac{2^{k-7}}{3}(38(k-3)+9(k-3)^2+(k-3)^3)\\ -\frac{2^{k-7}}{3}(38(k-3)+9(k-3)^2+(k-3)^3)\\  a_{k-4} = \frac{2^{k-11}}{3}(378+179(k-4)+18(k-4)^2+(k-4)^3) \\  a_{k-5} = \frac{2^{k-13}}{15}(9864(k-5)+3030(k-5)^2+515(k-5)^3+30(k-5)^4+(k-5)^5) -\frac{2^{k-13}}{15}(9864(k-5)+3030(k-5)^2+515(k-5)^3+30(k-5)^4+(k-5)^5)  \\ a_{k-6} = \frac{2\cdot2^{n-17}}{45}(125640(k-6)+90634(k-6)^2+12915(k-6)^3+1165(k-6)^4+45(k-6)^5+(k-6)^6)  \end{equation}  The we can see that the power of $2$ always seems to contain the next prime, although $a_{k-6}$ is somewhat forced.