## how to find consecutive nonprimes

We use a kind of reverse Eratosthenes Sieve. Observe that n! = 2·3·4·...·n is divisible by every integer from 2 to n. It follows that each of the n integers (n + 1)! + 2, (n + 1)! + 3, ..., (n + 1)! + n, (n + 1)! + (n + 1) is nonprime.

This method yields the following results:

n = 2: (3! + 2, 3! + 3) = (8, 9)

n = 3: (4! + 2, 4! + 3, 4! + 4) = (26, 27, 28)

n = 4: (5! + 2, 5! + 3, 5! + 4, 5! + 5) = (122, 123, 124, 125)

n = 5: (6! + 2, 6! + 3, 6! + 4, 6! + 5, 6! + 6) = (722, 723, 724, 725, 726)

n = 6: (7! + 2, 7! + 3, 7! + 4, 7! + 5, 7! + 6, 7! + 7) = (5042, 5043, 5044, 5045, 5046, 5047)

To find 1,000 consecutive nonprimes, simply consider the interval from 1,001! + 2 to 1,001! + 1,001.

Note: We do not claim that the above method provides us with the first such sequence. It clearly does not for small values of n such as 3 (8, 9, 10) and 4 (24 thru 27).