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Benedict Irwin edited untitled.tex
about 8 years ago
Commit id: 3d8e67059382296cef8b29252435a8ee679fd56e
deletions | additions
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index dbe4881..fcd5b11 100644
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we can see the subsequence $71,213,19170,19170,1811565$ is $1,3,270,270,25515$ when divided by $71$.
The number of $3$'s in the lead coeficcient are $1,3,3,5,6,8$ which could be a number of sequences.
The number of $3$'s in the second coefficients are $0,1,3,3,6$ which could also be many sequences.
Then we have
\begin{equation}
\pi_m(k) = \sum_{i=0}^{m-1}\frac{\left[\prod_{j=0}^{i} (m-j)\right]}{d_{i+1}}\sigma_i(m)k^{m-i}
\end{equation}
and we focus on the $\sigma_i(m)$ polynomials \begin{equation}
\simga_0(m)=3 \\
\sigma_1(m)=27m+71 \\
\sigma_2(m)=27m^2+213m-528\\
\sigma_3(m)=1215m^3+19170 m^2-69835 m+191522\\
\sigma_4(m)=729m^4+19170m^3-66945m^2+199686 m-1863360\\
\sigma_5(m)=45927 m^5+1811565 m^4-3671325 m^3-20584781 m^2-243156858 m-181415824
\end{equation}