The list is 1 2 4 8 16 26 42 64 93. A number in this list is wrong.
Finite differences propagation error: \begin{array}{|c|c|c|c|} \hline f(x)& \Delta^1{f}& \Delta^2{f}& \Delta^3{f} \\ \hline f_{0}& \Delta^1{f_{0}}& \Delta^2{f_{0}}& \Delta^3{f_{0}}+\epsilon \\ \hline f_{1}& \Delta^1{f_{1}}& \Delta^2{f_{1}}+\epsilon& \Delta^3{f_{1}}-3\epsilon \\ \hline f_{2}& \Delta^1{f_{2}}+\epsilon& \Delta^2{f_{2}}-2\epsilon& \Delta^3{f_{2}}+3\epsilon \\ \hline f_{3}+\epsilon& \Delta^1{f_{3}}-\epsilon& \Delta^2{f_{3}}+\epsilon& \Delta^3{f_{3}}-\epsilon \\ \hline f_{4}& \Delta^1{f_{4}}& \Delta^2{f_{4}}& - \\ \hline f_{5}& \Delta^1{f_{5}}& -& - \\ \hline f_{6}& -& -& - \\ \hline \end{array} And so on.
\begin{array}{|c|c|c|c|} \hline f(x)& \Delta^1{f}& \Delta^2{f}& \Delta^3{f}& \Delta^4{f}& \Delta^5{f}& \Delta^6{f}& \Delta^7{f}& \Delta^8{f} \\ \hline 1& 1& 1& 1& 1& -5& 15& -35& 70 \\ \hline 2& 2& 2& 2& -4& 10& -20& 35& -\\ \hline 4& 4& 4& -2& 6& -10& 15& -& -\\ \hline 8& 8& 2& 4& -4& 5& -& -& -\\ \hline 16& 10& 6& 0& 1& -& -& -& - \\ \hline 26& 16& 6& 1& -& -& -& -& - \\ \hline 42& 22& 7& -& -& -& -& -& - \\ \hline 64& 29& -& -& -& -& -& -& - \\ \hline 93& -& -& -& -& -& -& - \\ \hline \end{array}
$How\ to\ find\ \Delta^1{f_{0}}=f_{1}-f_{0}$
$\Delta^1{f_{1}}=f_{2}-f_{1} and\ so\ on\ $
$\Delta^2{f_{0}}=\Delta^1{f_{1}}-\Delta^1{f_{0}}$
$\Delta^2{f_{1}}=\Delta^1{f_{2}}-\Delta^1{f_{1}} and\ so\ on\ $
The column $1,-4,6,-4,1$ looks like Pascal's triangle with alternate terms reversed. If you work diagonally downward from the $6$ (because you portray the differences diagonally) you find the term to fix. If you centered the differences under where they come from it would be right above. Change the $16$ to $15$ and you have a cubic.