Smallest 10 Primes in Arithmetic Sequence

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Theorem

The smallest $10$ primes in arithmetic sequence are:

$199 + 210 n$

for $n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.

These are also the smallest $8$ and $9$ primes in arithmetic sequence.


Proof

\(\ds 199 + 0 \times 210\) \(=\) \(\ds 199\) which is the $46$th prime
\(\ds 199 + 1 \times 210\) \(=\) \(\ds 409\) which is the $80$th prime
\(\ds 199 + 2 \times 210\) \(=\) \(\ds 619\) which is the $114$th prime
\(\ds 199 + 3 \times 210\) \(=\) \(\ds 829\) which is the $145$th prime
\(\ds 199 + 4 \times 210\) \(=\) \(\ds 1039\) which is the $175$th prime
\(\ds 199 + 5 \times 210\) \(=\) \(\ds 1249\) which is the $204$th prime
\(\ds 199 + 6 \times 210\) \(=\) \(\ds 1459\) which is the $232$nd prime
\(\ds 199 + 7 \times 210\) \(=\) \(\ds 1669\) which is the $263$rd prime
\(\ds 199 + 8 \times 210\) \(=\) \(\ds 1879\) which is the $289$th prime
\(\ds 199 + 9 \times 210\) \(=\) \(\ds 2089\) which is the $316$th prime

This sequence is A033168 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).


But note that $199 + 10 \times 210 = 2299 = 11^2 \times 19$ and so is not prime.

Now we show that this is the smallest $10$ primes in arithmetic sequence.


By Divisibility of Common Difference of Arithmetic Sequence of Primes, the common difference $d$ of any $10$ primes in arithmetic sequence must be divisible by all primes less that $10$.

That is, the common difference is a multiple of $2 \times 3 \times 5 \times 7 = 210$.


Suppose $d = 210$.

Then the first term $p_1$ of the arithmetic sequence cannot be $11$, as $11 + 210 = 221 = 13 \times 17$.

Note that $210 \equiv 1 \pmod {11}$.

Hence if the $p_1$ gives a remainder $r > 1$ upon divison by $11$, the $\paren {12 - r}^{th}$ term of the sequence is equivalent to:

$r + \paren {12 - r - 1} \times 210 \equiv 0 \pmod {11}$

and is divisible by $11$, which is impossible as all terms are primes.

Therefore $p_1 \equiv 1 \pmod {11}$.

The first few primes of this form are:

$23, 67, 89, 199, \dots$

and we eliminate the first three candidates because:

$23 + 5 \times 210 = 1073 = 29 \times 37$
$67 + 3 \times 210 = 697 = 17 \times 41$
$89 + 1 \times 210 = 229 = 13 \times 23$

which are not primes, and we have checked that $199 + 210 n$ are indeed $10$ primes in arithmetic sequence.


Now suppose $d > 210$.

Then $d \ge 420$ and the last term is greater than $420 \times 9 = 3780$.

This shows that the one we found here is the smallest $10$ primes in arithmetic sequence.

$\Box$


Smallest $9$ primes in arithmetic sequence

Most of the analysis above applies, but we need to also consider $p_1 \equiv 2 \pmod {11}$.

This gives the first few primes:

$13, 79, 101, 167, 211, \dots$

and we have:

$13 + 6 \times 210 = 1273 = 19 \times 67$
$79 + 1 \times 210 = 289 = 17^2$
$101 + 3 \times 210 = 731 = 17 \times 43$
$167 + 1 \times 210 = 377 = 13 \times 29$
$221 > 199$

so there are no smaller $9$ primes in arithmetic sequence.

$\Box$


Smallest $8$ primes in arithmetic sequence

Most of the analysis above applies, but we need to also consider $p_1 \equiv 3 \pmod {11}$.

This gives the first few primes:

$3, 47, 113, 157, 179, 223, \dots$

and we have:

$3 + 1 \times 210 = 213 = 3 \times 71$
$47 + 7 \times 210 = 1517 = 37 \times 41$
$113 + 1 \times 210 = 323 = 17 * 19$
$157 + 5 \times 210 = 1207 = 17 * 71$
$179 + 7 \times 210 = 1649 = 17 * 97$
$223 > 199$

so there are no smaller $8$ primes in arithmetic sequence.

$\blacksquare$


Sources

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