Calculate the series 5,7,17,47,115,?

Fill the series 5,7,17,47,115,? find the value in the "?".

The function is:

$\small{ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } \\\\ a_n &=& d_0 + (n-1)d_1 \\ &+& \frac12 (n-1)(n-2)d_2\\ &+& \frac13\cdot \frac12 (n-1)(n-2)(n-3)d_3 \\ &+& \frac14\cdot \frac13 \cdot \frac12 (n-1)(n-2)(n-3)(n-4)d_4 \\\\ a_n &=&d_0+ (n-1) \{d_1+(n-2)[\frac12 d_2+(n-3)( \frac16 d_3 + \frac{1}{24}(n-4)d_4) ] \} \\ \dots \\ a_n &=& d_0 -d_1+d_2-d_3+d_4 \\ &+& n\cdot \left( \frac{12d_1-18d_2+22d_3-25d_4}{12} \right) \\ &+& n^2\cdot \left( \frac{12d_2-24d_3+35d_4}{24} \right) \\ &+& n^3\cdot \left( \frac{4d_3-10d_4}{24} \right) \\ &+& n^4\cdot \left( \frac{d_4}{24} \right) \\\\ \end{array} }$

we have:

$d_0 = 5\\ d_1 = 2\\ d_2 = 8\\ d_3 = 12\\ d_4 = 6$

$\begin{array}{rcl} a_n &=& d_0 -d_1+d_2-d_3+d_4 \\ &+& n\cdot \left( \frac{12d_1-18d_2+22d_3-25d_4}{12} \right) \\ &+& n^2\cdot \left( \frac{12d_2-24d_3+35d_4}{24} \right) \\ &+& n^3\cdot \left( \frac{4d_3-10d_4}{24} \right) \\ &+& n^4\cdot \left( \frac{d_4}{24} \right) \\\\ a_n &=& 5-2+8-12+6\\ &+& n\cdot \left( \frac{12\cdot 2-18\cdot 8+22\cdot 12-25\cdot 6}{12} \right) \\ &+& n^2\cdot \left( \frac{12\cdot 8-24\cdot 12+35\cdot 6}{24} \right) \\ &+& n^3\cdot \left( \frac{4\cdot 12-10\cdot 6}{24} \right) \\ &+& n^4\cdot \left( \frac{6}{24} \right) \\\\ a_n &=& 5\\ &-& \frac12 \cdot n\\ &+& \frac34 \cdot n^2 \\ &-& \frac12 \cdot n^3 \\ &+& \frac14 \cdot n^4\\\\ a_n &=& 5 - \frac12 \cdot n + \frac34 \cdot n^2 - \frac12 \cdot n^3 + \frac14 \cdot n^4 \end{array}$

One possibility is 245:

.

Very impressive but how did you work it out Alan?  

It wasn't even in the oeis ....

Fill the series 5, 7, 17, 47, 115, ?  find the value in the "?".

$\small{ \begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 5} && 7 && 17 && 47 && 115 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 2} && 10 && 30 && 68 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 8} && 20 && 38 && \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 12} && 18 && \cdots \\ \text{4. Difference } &&&&& {\color{red}d_4 = 6} && \cdots \\ \end{array} }$

$\boxed{~ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } \end{array} ~} \\\\ $

$\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red} 5 } + \binom{n-1}{1}\cdot {\color{red} 2 } + \binom{n-1}{2}\cdot {\color{red} 8 } + \binom{n-1}{3}\cdot {\color{red} 12 } + \binom{n-1}{4}\cdot {\color{red} 6 } \\\\ a_{6} &=& \binom{5}{0}\cdot {\color{red} 5 } + \binom{5}{1}\cdot {\color{red} 2 } + \binom{5}{2}\cdot {\color{red} 8 } + \binom{5}{3}\cdot {\color{red} 12 } + \binom{5}{4}\cdot {\color{red} 6 } \\\\ a_{6} &=& 1\cdot {\color{red} 5 } + 5\cdot {\color{red} 2 } + 10\cdot {\color{red} 8 } + 10\cdot {\color{red} 12 } + 5\cdot {\color{red} 6 } \\\\ a_{6} &=& 5+10+80+120+30 \\\\ \mathbf{a_{6}} & \mathbf{=}& \mathbf{245} \end{array}$

I just fitted a 4th order polynomial to the sequence.  $f(n)=\Sigma_{k=0}^4 a_kn^k$

There are five unknowns (the a_k) and five numbers in the sequence.

5            7               17             47               115                 X  (245)

      +2            +10           +30             +68               +130

             +8               +20             +38           +62

                     +12               +18              +24

                               +6                 +6

18+6=24

38+24=62

68+62=130

115+130=245