Arithmetic Sequence of 16 Primes

Theorem

The $16$ integers in arithmetic sequence defined as:

$2\,236\,133\,941 + 223\,092\,870 n$

are prime for $n = 0, 1, \ldots, 15$.

Proof

First we note that:

$2\,236\,133\,941 - 223\,092\,870 = 2\,013\,041\,071 = 53 \times 89 \times 426\,763$

and so this arithmetic sequence of primes does not extend to $n < 0$.

 $\ds 2\,236\,133\,941 + 0 \times 223\,092\,870$ $=$ $\ds 2\,236\,133\,941$ which is prime $\ds 2\,236\,133\,941 + 1 \times 223\,092\,870$ $=$ $\ds 2\,459\,226\,811$ which is prime $\ds 2\,236\,133\,941 + 2 \times 223\,092\,870$ $=$ $\ds 2\,682\,319\,681$ which is prime $\ds 2\,236\,133\,941 + 3 \times 223\,092\,870$ $=$ $\ds 2\,905\,412\,551$ which is prime $\ds 2\,236\,133\,941 + 4 \times 223\,092\,870$ $=$ $\ds 3\,128\,505\,421$ which is prime $\ds 2\,236\,133\,941 + 5 \times 223\,092\,870$ $=$ $\ds 3\,351\,598\,291$ which is prime $\ds 2\,236\,133\,941 + 6 \times 223\,092\,870$ $=$ $\ds 3\,574\,691\,161$ which is prime $\ds 2\,236\,133\,941 + 7 \times 223\,092\,870$ $=$ $\ds 3\,797\,784\,031$ which is prime $\ds 2\,236\,133\,941 + 8 \times 223\,092\,870$ $=$ $\ds 4\,020\,876\,901$ which is prime $\ds 2\,236\,133\,941 + 9 \times 223\,092\,870$ $=$ $\ds 4\,243\,969\,771$ which is prime $\ds 2\,236\,133\,941 + 10 \times 223\,092\,870$ $=$ $\ds 4\,467\,062\,641$ which is prime $\ds 2\,236\,133\,941 + 11 \times 223\,092\,870$ $=$ $\ds 4\,690\,155\,511$ which is prime $\ds 2\,236\,133\,941 + 12 \times 223\,092\,870$ $=$ $\ds 4\,913\,248\,381$ which is prime $\ds 2\,236\,133\,941 + 13 \times 223\,092\,870$ $=$ $\ds 5\,136\,341\,251$ which is prime $\ds 2\,236\,133\,941 + 14 \times 223\,092\,870$ $=$ $\ds 5\,359\,434\,121$ which is prime $\ds 2\,236\,133\,941 + 15 \times 223\,092\,870$ $=$ $\ds 5\,582\,526\,991$ which is prime

But note that $2\,236\,133\,941 + 16 \times 223\,092\,870 = 5\,805\,619\,861 = 79 \times 73\,488\,859$ and so is not prime.

$\blacksquare$