Photoelectric Primes 2:The rise of fallen

Apply the Method of Differences .

$\begin{array}{l|c|r}n& f(n)& D_1& D_2& D_3& D_4\\\hline 1& 2& 1& 1& -1& 3\\\hline 2& 3& 2& 0& 2\\\hline 3& 5& 2& 2\\\hline 4& 7& 4\\\hline 5& 11\\\hline 6\end{array}$

$\begin{array}{l|c|r}n& f(n)& D_1& D_2& D_3& D_4\\\hline 1& 2& 1& 1& -1& 3\\\hline 2& 3& 2& 0& 2& 3\\\hline 3& 5& 2& 2\\\hline 4& 7& 4\\\hline 5& 11\\\hline 6\end{array}$

$\begin{array}{l|c|r}n& f(n)& D_1& D_2& D_3& D_4\\\hline 1& 2& 1& 1& -1& 3\\\hline 2& 3& 2& 0& 2& 3\\\hline 3& 5& 2& 2& 5\\\hline 4& 7& 4& 7\\\hline 5& 11& 11\\\hline 6& \boxed{22}\end{array}$

A quartic polynomial is of the form $ax^4+bx^3+cx^2+dx+e=f(x)$ . If we substitute in the given values, a linear system results: $\begin{cases} {a+b+c+d+e = 2} \\ {16a+8b+4c+2d+e = 3} \\ {81a+27b+9c+3d+e = 5} \\ {256a+64b+16c+4d+e = 7} \\ {625a+125b+25c+5d+e = 11} \end{cases}$

Put into $5\text{x}6$ matrix form, this system is written as $\left[ \begin{array}{cccccc} 1 & 1 & 1 & 1 & 1 & 2 \\ 16 & 8 & 4 & 2 & 1 & 3 \\ 81 & 27 & 9 & 3 & 1 & 5 \\ 256 & 64 & 16 & 4 & 1 & 7 \\ 625 & 125 & 25 & 5 & 1 & 11 \end{array} \right]$

and reduces to $\left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0.125 \\ 0 & 1 & 0 & 0 & 0 & -1.41\overline{6} \\ 0 & 0 & 1 & 0 & 0 & 5.875 \\ 0 & 0 & 0 & 1 & 0 & -8.58\overline{3} \\ 0 & 0 & 0 & 0 & 1 & 6 \end{array} \right]$

$0.125\cdot6^4-1.41\overline{6}\cdot6^3+5.875\cdot6^2-8.58\overline{3}\cdot6+6=\boxed{22}$

A simple application of linear algebra.