USING: kernel math math.primes project-euler.common sequences ;
IN: project-euler.027
-! https://projecteuler.net/index.php?section=problems&id=27
+! https://projecteuler.net/problem=27
! DESCRIPTION
! -----------
! 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.
+! 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.
-! Using computers, 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.
+! Using computers, 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:
! 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.
+! 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.
! SOLUTION