-
Notifications
You must be signed in to change notification settings - Fork 0
/
p027.py
46 lines (32 loc) · 1.2 KB
/
p027.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 26 12:49:00 2020
@author: zhixia liu
"""
"""
Project Euler 27: Quadratic primes
Euler discovered the remarkable quadratic formula:
n2+n+41
It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤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 0≤n≤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.
"""
#%% naive
from sympy import isprime,primerange
primel = primerange(1,1000)
maxl = 0
p=0
for b in primel:
for a in range(-b,1000):
n = 0
while isprime(n**2 + a*n + b):
n += 1
if n>maxl:
p = a*b
maxl = n
print(a,b,n)
print(p)