Problem 2342. Numbers spiral diagonals (Part 2)
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Interesting problem! If I'm not mistaken, your description of the problem is not entirely accurate, however. For n = 1, the spiral matrix is just [ 1 ], for which the share of primes on the main diagonals is zero, below any given 0 < p < 1; so strictly speaking the correct answer to the problem as posed would be 1 for any p.
Nice observation, @Christian!
I have edited the question statement, it (now) asks for an odd integer greater than 1.
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You added a few primes in each iteration... Very nice!
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