Symbol(clack:cancel)

solutions/0000_0029/0027.QuadraticPrimes/euler_27.py

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+"""
+Euler discovered the remarkable quadratic formula: n^2 + n + 41
+
+It turns out that the formula will produce 40 primes for the consecutive integer values 0<=n<=39.
+However, when n = 40, 40^2+40+41=40*(40+1)+41 is divisible by 41, and certainly when n = 41,
+41^2+41+41 is clearly divisible by 41.
+
+The incredible formula n^2-79n+1601 was discovered, which produces 80 primes for the consecutive values
+0<=n<=79 . The product of the coefficients, -79 and 1601, is -126479 .
+
+Considering quadratics of the form:
+n^2+an+b , where |a|<1000 and |b|<=1000
+where |n| is the modulus/absolute value of n
+e.g. |11| = 11 and |-4| = 4
+
+Find the product of the coefficients, a and b, for the quadratic expression that produces
+the maximum number of primes for consecutive values of n, starting with n=0.
+"""
+
+from typing import Callable
+
+
+def quadratic_primes(a: int, b: int) -> Callable[[int], int]:
+    return lambda n: n**2 + a * n + b
+
+
+def is_prime(n: int) -> bool:
+    if n < 2:
+        return False
+    if n == 2:
+        return True
+    if n % 2 == 0:
+        return False
+    for i in range(3, int(n**0.5) + 1, 2):
+        if n % i == 0:
+            return False
+    return True
+
+
+def max_prime_quadratic(a_max: int, b_max: int) -> tuple[int, int, int]:
+    max_primes = 0
+    max_a = 0
+    max_b = 0
+    for a in range(-a_max, a_max + 1):
+        for b in range(-b_max, b_max + 1):
+            primes = 0
+            n = 0
+            while is_prime(quadratic_primes(a, b)(n)):
+                primes += 1
+                n += 1
+            if primes > max_primes:
+                max_primes = primes
+                max_a = a
+                max_b = b
+    return max_a, max_b, max_a * max_b
+
+
+def main():
+    a_max = 1000
+    b_max = 1000
+    max_a, max_b, product = max_prime_quadratic(a_max, b_max)
+    print(f"{max_a} x {max_b} = {product}.")
+
+
+if __name__ == "__main__":
+    main()