Ok so this solution is supersuperslow but it worked

27/main.py

@@ -4,4 +4,66 @@
 # n^2 + an + b, where abs(a) < 1000 and abs(b) <= 1000
 
 # Find the product of the coefficients, a * b, for the quadratic expression that
-# produces the maximum number of primes for consecutive values of n, starting from n = 0
+# produces the maximum number of primes for consecutive values of n, starting from n = 0
+
+import numpy as np
+import math
+
+def quad(a, b, n):
+	return n**2 + a * n + b
+
+# Thing that immediately stands out is that b always needs to be prime itself,
+# since for n=0 only b remains, so that thins the pool a lot
+# So get the prime thingy out again
+
+def getPrimes(n):
+	primes = [1 for p in range(n)]
+	primes[0] = 0
+	primes[1] = 0
+
+	x = 2
+	while x < np.sqrt(n):
+		if primes[x] == 1:
+
+			for i in range(x+x, n, x):
+				primes[i] = 0
+
+		x += 1
+
+	result = []
+	for i in range(n):
+		if primes[i]:
+			result.append(i)
+
+	return result
+
+def getPrimeSequenceLength(a, b, primes):
+	n = 0
+	while True:
+		if quad(a, b, n) in primes:
+			n += 1
+		else:
+			break
+
+	return n - 1
+
+def main():
+	print("Hello, this is Patrick")
+
+	ps = getPrimes(1000000)
+	littleps = list(filter(lambda p: p < 1000, ps))
+
+	maxprod = 0
+	maxlength = 0
+	for b in littleps:
+		for a in range(-b, 1000):
+			psl = getPrimeSequenceLength(a, b, ps)
+			if psl > maxlength:
+				maxlength = psl
+				maxprod = a * b
+
+	print(f"The maximum prime sequence {maxlength} was found for {maxprod}")
+	
+
+if __name__ == "__main__":
+	main()