Project Euler Problem 27 Statement
Euler published 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.
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.
Note: The quadratic formula needs only to generate prime numbers, not necessarily unique nor consecutive prime numbers.
This problem is wanting us to find two coefficients, a and b, for a quadratic expression,
Using a simple iteration of a and b over their range will produce a result for this specific problem, but takes too long for the more complex inquiry from HackerRank.
There are a few observations that will help improve performance.
b has to be prime
The problem requires our n to start at 0, and since we are looking for prime numbers with consecutive values of n, our first result from a quadratic expression would be 02 – 0 + b, which evaluates simply to b. So b has to be positive, prime and less than 1000 (for this problem), which helps reduce the search space significantly.
Additionally, we can now start
n at 1 since we know the first outcome is guaranteed prime. Our sieve,
prime_sieve(L), generates primes less than
L. Adding 1 to our limit, guarantees
L gets included should it be a prime number.
The prime number 2 is excluded from the possible starting primes as it evaluates to an even number after 1 or 2 iterations.
1-a+b must be odd and possibly prime
If you further consider when n=1, then the expression becomes 1-a+b or 1+a+b, which must always be evaluated to an odd number to yield a possible prime. We know b will always be an odd prime so 1+b will be even. We can now infer that a must be odd and |a| < b.
HackerRank's Project Euler 27 adds the constraint that the coefficients a and b are in the range [-2000,-42] U [42,2000]. The positive constraint on a and b can be ignored, as we previously established, effectively halving the search range. We are asked to print the two coefficients a and b instead of the product of a and b as the original problem required.
HackerRank simply changes the 1000 limit to any N, where 42 ≤ N ≤ 2000.
Python Source Code
from Euler import prime_sieve, is_prime L = int(input()) nmax = 0 for b in prime_sieve(L+1)[1:]: #start with prime number 3 for a in range(-b+2, 0, 2): n = 1 while is_prime(n*n + a*n + b): n += 1 if n>nmax: nmax, p = n, (a,b) print (p,p)
- Look at the repl.it for the