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Highest Number of Consecutive Primes

Sam Benry
Ranch Hand

Joined: Mar 21, 2008
Posts: 89
Euler published the remarkable quadratic formula:

n� + 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, 41� + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula n� − 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:

n� + 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.

This is weird, the code I wrote is very simple, it gives correct answers for a = 1 b = 41 and a=-79 b = 1601
but when I run it to find the largest number of consecutive primes, I get 1011 consecutive prime numbers when a = -999 and b = 61
making the product -60939
but this is turning out to be the wrong answer...
anyone can find me an error in the code?
Jim Yingst

Joined: Jan 30, 2000
Posts: 18671
Well, look at the "prime" numbers generated by the formula when a = -999 and b = 61. What do you get when n = 0? n = 1? n = 2? Are all those numbers prime? If not, how can you modify the isPrime() method to correctly identify whether a number is prime or not?

"I'm not back." - Bill Harding, Twister
I agree. Here's the link:
subject: Highest Number of Consecutive Primes
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