A polynomial P(x) of degree 3n has the value 2 at x= 0, 3, 6,..., 3n, and:
the value 1 at x= 1, 4, 7, ... , 3n-2, and:
the value 0 at x= 2, 5, 8, ... , 3n-1.

Given that P(3n+1) = 730, find n.

n = 4

...2  1   0   2   1   0   2   1   0   2  1  0  2  730

  ..-1  -1  2  -1  -1   2  -1   -1  2  -1 -1  2  728
    ...0  3  -3  0   3   -3   0   3  -3  0  3  726
      ...3 -6  3   3  -6    3   3   -6  3  3  723
        ...-9  9  0  -9    9   0   -9  9  0  720
         ...18 -9  -9  18  -9  -9   18 -9  720
           ...-27  0  27 -27  0  27  -27  729
              ...27 27 -54  27  27 -54  756
                ...0  -81  81  0  -81  810
                 ...-81 162 -81 -81  891
                   ...243 -243  0  972
                     ...-486  243 972
                      ...729  729
                          ...0

The top row shows the given values of P(x), at unit intervals, the
last value being P(3n+1). The rows below show the first differences
to the 13^th differences. Dots to the left indicate that we don’t yet
know the value of n and therefore how many three-cycles to
include in each set.
However, the zero that appears as a thirteenth difference, is the
first one to be at the tip of a triangle that extends upwards and
is derived from a set of P(x) values including 730.
A 13^th difference being zero shows that a polynomial of degree 12
can pass through the 14 points represented on the top row,
so 3n = 12, giving n = 4.

Perhaps there are bigger values of n that will work..?

Posted by Harry on 2016-05-08 14:58:49