+26388 1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?
\(\small{ \begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1} && 6 && 20 && 51 && 189 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 14 && 31 && 138 && 328 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 9} && 17 && 107 && 190 && 1352 && \cdots \\ \text{3. Difference } {\color{red}d_3 = 8} \\ \text{4. Difference } {\color{red}d_4 = 82} \\ \text{5. Difference } {\color{red}d_5 = -89} \\ \text{6. Difference } {\color{red}d_6 = 1175} \\ \text{7. Difference } {\color{red}d_7 = -4908} \\ \text{8. Difference } {\color{red}d_8 = 19363} \\ \end{array} }\)
\(\small{ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ &+& \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } + \binom{n-1}{5}\cdot {\color{red}d_5 }\\ &+& \binom{n-1}{6}\cdot {\color{red}d_6 } + \binom{n-1}{7}\cdot {\color{red}d_7 } + \binom{n-1}{8}\cdot {\color{red}d_8 }\\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 1 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 9 } \\ &+& \binom{n-1}{3}\cdot {\color{red} 8 } + \binom{n-1}{4}\cdot {\color{red} 82 } + \binom{n-1}{5}\cdot {\color{red} (-89) } \\ &+& \binom{n-1}{6}\cdot {\color{red} 1175 } + \binom{n-1}{7}\cdot {\color{red} (-4908) } + \binom{n-1}{8}\cdot {\color{red} 19363 } \\\\ a_{10} &=& \binom{9}{0}\cdot {\color{red} 1 } + \binom{9}{1}\cdot {\color{red} 5 } + \binom{9}{2}\cdot {\color{red} 9 } \\ &+& \binom{9}{3}\cdot {\color{red} 8 } + \binom{9}{4}\cdot {\color{red} 82 } + \binom{9}{5}\cdot {\color{red} (-89) } \\ &+& \binom{9}{6}\cdot {\color{red} 1175 } + \binom{9}{7}\cdot {\color{red} (-4908) } + \binom{9}{8}\cdot {\color{red} 19363 } \\ a_{10} &=& 1\cdot {\color{red} 1 } + 9\cdot {\color{red} 5 } + 36\cdot {\color{red} 9 } \\ &+& 84\cdot {\color{red} 8 } + 126\cdot {\color{red} 82 } + 126\cdot {\color{red} (-89) } \\ &+& 84\cdot {\color{red} 1175 } + 36\cdot {\color{red} (-4908) } + 9\cdot {\color{red} 19363 } \\ a_{10} &=& 1 + 45 + 324 +672 + 10332 -11214 +98700 - 176688+174267 \\ \mathbf{a_{10}} & \mathbf{=}& \mathbf{96439} \end{array} }\)
+118658 I have no idea, could you please check that all the terms are correct?
+118658 Well it is not in the OEIS so I do not like your chances of getting an answer :)
+33653 You can fit an order 8 polynomial to these numbers. The coefficients look awful but the next term given by the polynomial is 96439.
+26388 1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?
\(\small{ \begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 1} && 6 && 20 && 51 && 189 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 5} && 14 && 31 && 138 && 328 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 9} && 17 && 107 && 190 && 1352 && \cdots \\ \text{3. Difference } {\color{red}d_3 = 8} \\ \text{4. Difference } {\color{red}d_4 = 82} \\ \text{5. Difference } {\color{red}d_5 = -89} \\ \text{6. Difference } {\color{red}d_6 = 1175} \\ \text{7. Difference } {\color{red}d_7 = -4908} \\ \text{8. Difference } {\color{red}d_8 = 19363} \\ \end{array} }\)
\(\small{ \begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } \\ &+& \binom{n-1}{3}\cdot {\color{red}d_3 } + \binom{n-1}{4}\cdot {\color{red}d_4 } + \binom{n-1}{5}\cdot {\color{red}d_5 }\\ &+& \binom{n-1}{6}\cdot {\color{red}d_6 } + \binom{n-1}{7}\cdot {\color{red}d_7 } + \binom{n-1}{8}\cdot {\color{red}d_8 }\\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 1 } + \binom{n-1}{1}\cdot {\color{red} 5 } + \binom{n-1}{2}\cdot {\color{red} 9 } \\ &+& \binom{n-1}{3}\cdot {\color{red} 8 } + \binom{n-1}{4}\cdot {\color{red} 82 } + \binom{n-1}{5}\cdot {\color{red} (-89) } \\ &+& \binom{n-1}{6}\cdot {\color{red} 1175 } + \binom{n-1}{7}\cdot {\color{red} (-4908) } + \binom{n-1}{8}\cdot {\color{red} 19363 } \\\\ a_{10} &=& \binom{9}{0}\cdot {\color{red} 1 } + \binom{9}{1}\cdot {\color{red} 5 } + \binom{9}{2}\cdot {\color{red} 9 } \\ &+& \binom{9}{3}\cdot {\color{red} 8 } + \binom{9}{4}\cdot {\color{red} 82 } + \binom{9}{5}\cdot {\color{red} (-89) } \\ &+& \binom{9}{6}\cdot {\color{red} 1175 } + \binom{9}{7}\cdot {\color{red} (-4908) } + \binom{9}{8}\cdot {\color{red} 19363 } \\ a_{10} &=& 1\cdot {\color{red} 1 } + 9\cdot {\color{red} 5 } + 36\cdot {\color{red} 9 } \\ &+& 84\cdot {\color{red} 8 } + 126\cdot {\color{red} 82 } + 126\cdot {\color{red} (-89) } \\ &+& 84\cdot {\color{red} 1175 } + 36\cdot {\color{red} (-4908) } + 9\cdot {\color{red} 19363 } \\ a_{10} &=& 1 + 45 + 324 +672 + 10332 -11214 +98700 - 176688+174267 \\ \mathbf{a_{10}} & \mathbf{=}& \mathbf{96439} \end{array} }\)
+26388 1, 6, 20, 51, 189, 517, 2197, 4823, 14496, what comes next?
\(\small{ \begin{array}{rcr} a_1 &=& 1\\ a_2 &=& 6\\ a_3 &=& 20\\ a_4 &=& 51\\ a_5 &=& 189\\ a_6 &=& 517\\ a_7 &=& 2197\\ a_8 &=& 4823\\ a_9 &=& 14496\\ \end{array} }\)
\(\begin{array}{|cr|r|r|r|r|} \hline &1&20&189&2197&14496 \\ +&6&+51&+517&+4823&+\color{red}55555 \\ \hline =&7&7\underbrace{1}_{a_1}&70\underbrace{6}_{a_2}&70\underbrace{20}_{a_3}&700\underbrace{51}_{a_4} \\ \hline \end{array}\)
next comes \(\color{red}55555\)
+2 wow thanks guys ;D heureka and guest for helping me understand
conclusion:
1, 6, 20, 51, 189, 517, 2197, 4823, 14496, 55555
