Estimates of the number of magic squares, cubes, ... (hypercubes)
 Symmetrical and pandiagonal magic squares Incompatibilities and Contradictions

Symmetrical magic squares
These squares and the normal E-number are incompatible.

You can't use the magic series of simple magic squares in this case.
A symmetrical magic square only consists of symmetrical magic series and of series that do not contain complementary pairs of numbers.
With the numbers of these special series you would be able to estimate the number of symmetrical magic squares.
Dear reader, I am looking forward to your own results in this field.
See an example at the end of the page on multimagic squares.
Pandiagonal magic squares
There are 4(m - 1) unforced magic lines in a pandiagonal magic square of order m.
Order 5 is too small to apply statistic methods.

For order m = 6 the E-number equals 1.0 · 105.
But as you are a friend of magic squares you surely know that pandiagonal magic squares of order 6 do not exist. Frost and Planck proved that mathematically. We learn that we can't always trust the E-number. There may exist mathematical contradictions in special cases.

For m = 7 the E-number is 1.0 · 1015
In this case we also got an approximation by Monte Carlo Backtracking: N(7) = 1.2 · 1017
Thus the C-factor is very large: C = 120
This is caused by the high magicness probabilities of the last checked (broken) diagonals. We can assume that the C-factor increases with higher orders. But until today we didn't get any approximation function for the C-factor, because we have no approximations for m ≥ 8.
At least with C = 100 we get interesting lower borders for the numbers of pandiagonal magic squares:
 N(m) > 100 · E(m) Number of pandiagonal magic squares of order m ≠ 4k+2
N(6) = 0
N(7) > 1.0 · 1017
N(8) > 2.5 · 1031
N(9) > 3.2 · 1050
N(10) = 0
N(11) > 1 · 10104
N(12) > 7 · 10138
N(13) > 2 · 10179
N(14) = 0

Walter Trump, Nürnberg, Germany, (c) 2005-02-21 (last modified: 2005-03-04)