Euler Problem 58

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18  5  4  3 12 29
40 19  6  1  2 11 28
41 20  7  8  9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?


In [1]:
from sympy import isprime
corners = [3, 5, 7, 9]
squares = 5
primes = 3
delta = 8
sidelength = 3
while primes*10 >= squares:
    sidelength += 2
    for i in range(4):
        delta += 2
        corners[i] += delta
        primes += isprime(corners[i])
    squares += 4
print(sidelength)


26241

In [ ]: