There is one relatively simple optimization you can do, for a start, and it will likely make a big difference...

If you analyze the equation:

n² + an + b, where |a| < 1000 and |b| < 1000

you need to solve it for where n² + an + b is prime.

The first n value to use is 0. 0² +a0 +b is b, thus, in order to process any values, it is clear that b needs to be a prime number (and not negative either...).

Thus, you should rearrange your loops, and your outer loop should be on the first k prime numbers in your set, where the k'th prime is < 1000.

There are 168 primes less than 1000, so that will reduce your loops from 2000 'b' loops to just 168.

There is one relatively simple optimization you can do, for a start, and it will likely make a big difference...

If you analyze the equation:

n² + an + b, where |a| < 1000 and |b| < 1000

you need to solve it for where n² + an + b is prime.

The first n value to use is 0. 0² +a0 +b is b, thus, in order to process any values, it is clear that b needs to be a prime number (and not negative either...).

Thus, you should rearrange your loops, and your outer loop should be on the first k prime numbers in your set, where the k'th prime is < 1000.

There are 168 primes less than 1000, so that will reduce your loops from 2000 'b' loops to just 168.

Similarly, when you loop on only those 168 primes, you are guaranteed that the n == 0 result is prime, and you only need to test on the n > 0 condition.

Further, when n is 1, the equation becomes:

1 + a + b = prime

or, since you have primes already from your sieve, you can significantly restrict the data-set by doing:

a = prime - b - 1

for primes in the range of b - 998 through to b + 1000 (which will keep a within the limits of -999 to +999). Note that there are no primes less than 2, and only a few hundred primes less than the 'worst case' (b ==999) + 1000

Only if you satisfy those two conditions would I consider even calling the prime function f(a,b,n++)

There is one relatively simple optimization you can do, for a start, and it will likely make a big difference...

If you analyze the equation:

n² + an + b, where |a| < 1000 and |b| < 1000

you need to solve it for where n² + an + b is prime.

The first n value to use is 0. 0² +a0 +b is b, thus, in order to process any values, it is clear that b needs to be a prime number (and not negative either...).

Thus, you should rearrange your loops, and your outer loop should be on the first k prime numbers in your set, where the k'th prime is < 1000.

There is one relatively simple optimization you can do, for a start, and it will likely make a big difference...

If you analyze the equation:

n² + an + b, where |a| < 1000 and |b| < 1000

you need to solve it for where n² + an + b is prime.

The first n value to use is 0. 0² +a0 +b is b, thus, in order to process any values, it is clear that b needs to be a prime number (and not negative either...).

Thus, you should rearrange your loops, and your outer loop should be on the first k prime numbers in your set, where the k'th prime is < 1000.

There are 168 primes less than 1000, so that will reduce your loops from 2000 'b' loops to just 168.