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The following 5-by-5 squares of digits are not magic squares, but nevertheless they each have a crucial property which enabled me to generate them. What is that property? How could you generate more such squares?

3  3  5  8  9               9  5  4  8  7
4  5  1  4  7               0  1  7  2  8
6  6  3  1  9               4  8  2  8  2
2  2  9  5  3               5  2  8  1  0
6  4  8  3  2               2  3  5  3  6
$\endgroup$
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    $\begingroup$ Math.random()? :p $\endgroup$ – Novarg Mar 29 '15 at 0:26
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    $\begingroup$ @Novarg I don't write troll questions! :-p $\endgroup$ – Rand al'Thor Mar 29 '15 at 0:29
  • $\begingroup$ AND every row is zero, AND every column is zero. $\endgroup$ – Aravind Mar 29 '15 at 16:30
  • $\begingroup$ @Aravind what?? $\endgroup$ – Rand al'Thor Mar 29 '15 at 16:32
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    $\begingroup$ Write every number in binary. And AND. I realize that's not the intended solution, though. $\endgroup$ – Aravind Mar 29 '15 at 16:35
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The first one

writes the first 25 digits of $\pi$ in a spiral, beginning with 3 in the middle and moving off upwards and clockwise.

The second does the same for

$e$, beginning with 2 in the middle and moving off upwards and counter-clockwise.

The property is that they

represent irrational numbers in Ulam-type spirals.

To generate more,

pick other irrational numbers and do the same.

Here's one for

Feigenbaum's $\delta$ (4.669201609102990671853203...), starting off to the left:

3  0  2  3  5
9  0  6  1  8
1  6  4  0  1
0  6  9  2  7
2  9  9  0  6
$\endgroup$
  • $\begingroup$ Wow, I didn't even notice. Good job! $\endgroup$ – Joe Z. Apr 6 '15 at 9:31

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