Logic of numerical series

Question

One of our colleagues has written a numerical series to the whiteboard in our breakroom. Nobody until now could solve the logic behind this series:

1, 14, 3, -8, 27, 41, -91, 33, ...

Can anybody say how this series is continued and whats the logic to calculate it?

It's enough to drive one to despair.

Answer 1

Use Lagarange's Law(Method of finite difference) Write them like this: $$\begin{array}{c|c} 1&14&3&-8&27&41&-91&33&\Delta^0\\ (-)&1&14&3&-8&27&41&-91\\\hline ..&13&-11&-11&35&14&-132&124&\Delta^1\\ ...&(-)&13&-11&-11&35&14&-132\\\hline ...&...&-24&...&...&...&...&...&\Delta^2 \end{array}$$ Now you can write the series as : $$t_n=\binom {n-1}0\Delta^0 t_1+\binom {n-1}1\Delta^1 t_1+\binom {n-1}2\Delta^2 t_1+\cdots\\ =\sum_{k=0}^{n-1}\binom {n-1}k\Delta^k t_1+$$ _{which can be proved by mathematical induction Also take $\binom nr=0 ,\text{ for } r<0$}

where, $\Delta^rt_1$ is the first term of r^th difference, for e.g. $$\Delta^0t_1=1,\Delta^1t_1=13,\Delta^2t_1=-24\cdots$$

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_{Actually this method(as in here) is just a way of interpolation with the given data, if you use 8 terms in the formula you get eight terms correct, if for example a series of 2,4,6,8... ;the $\Delta^1 t_1=2$ and further all differences are zero so it correctly forms the function $2\times\binom {n-1}0+2\times\binom{n-1}1=2n$; this method can tell all terms if we had other terms too or until some level of difference became $0${or a special series}.}

Answer 2

Last week, I posted this answer:

You ask, "Can anybody say how this series is continued and what's the logic to calculate it?"

The answer is yes; the colleague who wrote it on the whiteboard can do both of those things.

That answer was deleted by a moderator.

Since then, no one here has been able to say how the series is continued, etc. (as OP correctly rejects anything based on Lagrange interpolation). So I think it's time to post a modified version of my deleted answer:

The answer is yes; only the colleague who wrote it on the whiteboard can do both of those things.