Question 1143335
<pre>
There will always an infinite number of polynomial sequences which
any subsequence of numbers will fit.  I hope your teacher knows that!

The first three numbers 

1, 4, 27, for the sequence are 1<sup>1</sup>, 2<sup>2</sup>, 3<sup>3</sup>,

so the fourth term should be 4<sup>4</sup> or 256, and the 5th term
should be 5<sup>5</sup> or 3125, but there is no way it can be the 3761,
which is a prime number.

So let's find one polynomial sequence it could be. We make this
difference table, where the first column is the given sequence.
Then immediately to the right of each number is that number
subtracted from what's immediately under it, (if there is a
number immediately under it.)

   1       3        20      70-X   3908-4x  
   4      23      X-50   3838-3X  
  27    X-27   3788-2X
   X  3761-X     
3761

So we let 3908-4x = 0
              -4x = -3908
                x = 977

Then that makes a nice looking sequence, since 977 is
about the proper looking size.

1, 4, 27, 977, 3761

That sequence has the nth term:

{{{a[n] = -54x^3+expr(4471/6)n^2-3111n+expr(9815/6)n-2509}}}

But there are an infinite number of other possibilities.
In fact you can pick any number for X and there will be
an infinite number of possibilities.  I suppose you are
supposed to find the simplest one.  How is that supposed
to be done?  If your teacher tell you how, please tell us, 
for we have no idea, although we have advanced degrees in
mathematics.

Edwin</pre>