```python
"""
Euler discovered the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. 
However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, 
and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

The incredible formula  n² − 79n + 1601 was discovered, which produces 80 primes for the 
consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479.

Considering quadratics of the form:

n² + 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.
"""
```


```python
import itertools
from sympy.ntheory import isprime

limit = 1000

abcombos = list(itertools.product(range(-(limit - 1),limit), repeat = 2))

n= 0
maxab = (0,0)
maxn = 0 

for a, b in abcombos: 
    n = 0      # start with n = 0
    while isprime((n+1)**2 + a*(n+1) + b): # while quadratic expression per (a,b) tuple produces primes: 
        n += 1 # increment result counter
    if n > maxn: # when the expression produces a non-prime for the first time, check if the previous run 
        maxn = n     # new n with the most primes per (a,b) tuple
        maxab = a, b # new (a,b) tuple 

print(maxab[0]*maxab[1])
```

    (-61, 971)
    70
    1601
    -59231



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ddc59e92d1ed089175f230404feb6bb27438bf6c
ipynb
Jupyter Notebook
Project_Euler-Problem_27-Quadratic_primes.ipynb
SmirnovAnton/project-euler-solutions
31d69fa2356b81395bf9267025da9d3d09e313cf
Apache-2.0
Project_Euler-Problem_27-Quadratic_primes.ipynb
SmirnovAnton/project-euler-solutions
31d69fa2356b81395bf9267025da9d3d09e313cf
Apache-2.0
Project_Euler-Problem_27-Quadratic_primes.ipynb
SmirnovAnton/project-euler-solutions
31d69fa2356b81395bf9267025da9d3d09e313cf
Apache-2.0
Qwen/Qwen-72B
1. YES
2. YES