Inspired by Project Euler n°28 and 58.
A n x n spiral matrix is obtained by starting with the number 1 and moving to the right in a clockwise direction.
For example with n=5, the spiral matrix is :
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
The sum of the numbers on the diagonals is 101 (See problem 2340) and you have 5 primes (3, 5, 7, 13, 17) out of the 9 numbers lying along both diagonals. So the prime ratio is 5/9 ≈ 55%.
With a 7x7 spiral matrix, the ratio is 62% (8 primes out of the 13 diagonal numbers).
What is the side length (always odd and greater than 1) of the square spiral for which the ratio of primes along both diagonals FIRST falls below p% ? (0<p<1)
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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.
This is a seriously good problem. Love it! You will have to optimize your code in order for the tests to not time out.
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