ordering of palindrome product space
prev palindrome prod order wrong; n = 999 exhausts python stack

The ordering I assumed only checked numbers on the diagonal and
superdiagonal matrix elements.  I need to check all the matrix elements:
(n - i)*(n - j) = n**2 - (i + j)*n + i*j.  However, my new, far simpler
idea is to assume that the largest palindrome in the set (or square
matrix with elements (n - i)*(n - j) for integers 0 <= i,j <= n) is much
closer to n**2 than to 0: n**2 > largest_base_ten_palindrome >> 0.
Now using 2-param estimator method: 0 << n - m < n

The assumption used is helpful to find the larges palindrome without
permuting the whole parameter space, however the palindrome itself can
be found in multiple binary combinations of the prime factors of the
palindrome.
just thought of a faster comparison method
completed 2 of 3 exercises as requested