A hard magic square puzzle

For a 3 x 3, there's only one distinct magic square; with a 4 x 4 there are 880 magic squares and for a 5th order-sized magic square, you can have 275,305,224 distinct magic squares.

Can you figure out a formula that can relate the number of members of a magic square with how many magic squares can exist for the third, fourth and fifth orders? That is:

9 1.................. third order level
16 880............. fourth order level
25 275,305,224.. fifth order level

where 9, 16 and 25 are the number of members for the third-, fourth- and fifth-order magic square respectively and 1, 880 and 275,305,224 are the number of distinct third-, fourth- and fifth-order magic squares respectively.

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Originally Posted by wonderboy1953

For a 3 x 3, there's only one distinct magic square; with a 4 x 4 there are 880 magic squares and for a 5th order-sized magic square, you can have 275,305,224 distinct magic squares.

Can you figure out a formula that can relate the number of members of a magic square with how many magic squares can exist for the third, fourth and fifth orders? That is:

9 1.................. third order level
16 880............. fourth order level
25 275,305,224.. fifth order level

where 9, 16 and 25 are the number of members for the third-, fourth- and fifth-order magic square respectively and 1, 880 and 275,305,224 are the number of distinct third-, fourth- and fifth-order magic squares respectively.

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Are you implying progress has been made beyond what is published on OEIS?

id:A006052 - OEIS Search Results

I calculated a(4) = 880 algorithmically in another thread of yours, haven't tried scaling it to a(5), might be best to translate into C++ for that for the sake of time, since my algorithm was I think suboptimal. Actually after reviewing that code I see it is rather hard-coded for 4x4 and would not easily be scaled for 5x5 (would require considerable reworking).

Are you implying progress has been made beyond what is published on OEIS? No, just the opposite.

Quote:

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Originally Posted by wonderboy1953

Are you implying progress has been made beyond what is published on OEIS? No, just the opposite.

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Maybe it's the wording of your question that strikes me; one would normally consider a formula neater and more efficient than an algorithm, such that knowing a formula would allow one to calculate a(6). Thus it seemed you were asking if anyone could solve an open problem in mathematics, but not stating that it is an open problem. Similarly to if someone were to ask if someone could solve the Goldbach conjecture in the Puzzles subforum, but stating as if it were any other puzzle with known solution (and of course, not mentioning the word conjecture anywhere in the post).

Quote:

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Originally Posted by undefined

Maybe it's the wording of your question that strikes me; one would normally consider a formula neater and more efficient than an algorithm, such that knowing a formula would allow one to calculate a(6). Thus it seemed you were asking if anyone could solve an open problem in mathematics, but not stating that it is an open problem. Similarly to if someone were to ask if someone could solve the Goldbach conjecture in the Puzzles subforum, but stating as if it were any other puzzle with known solution (and of course, not mentioning the word conjecture anywhere in the post).

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Let's say I believe no one has solved it yet (if it were solved I would have entered it into the math challenge subsubforum).

Quote:

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Originally Posted by wonderboy1953

Let's say I believe no one has solved it yet (if it were solved I would have entered it into the math challenge subsubforum).

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Fair enough, a couple of related thoughts

1) If you let people know it's a known and open problem, it could save people some time researching, etc.
2) If you don't let people know it's an open problem, you could get the "good will hunting" effect :)

Maybe I should not have brought it up, heh