Longest Sequence of Consecutive Primes in Arithmetic Sequence
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Sequence
The longest known sequence of consecutive prime numbers in arithmetic sequence starts at $121 \, 174 \, 811$, has a length of $6$ and a common difference of $30$:
- $121 \, 174 \, 811, 121 \, 174 \, 841, 121 \, 174 \, 871, 121 \, 174 \, 901, 121 \, 174 \, 931, 121 \, 174 \, 961$
This sequence is A081734 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
\(\ds 121 \, 174 \, 811 + 0 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 811\) | which is the $6 \, 904 \, 737$th prime | |||||||||||
\(\ds 121 \, 174 \, 811 + 1 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 841\) | which is the $6 \, 904 \, 738$th prime | |||||||||||
\(\ds 121 \, 174 \, 811 + 2 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 871\) | which is the $6 \, 904 \, 739$th prime | |||||||||||
\(\ds 121 \, 174 \, 811 + 3 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 901\) | which is the $6 \, 904 \, 740$th prime | |||||||||||
\(\ds 121 \, 174 \, 811 + 4 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 931\) | which is the $6 \, 904 \, 741$st prime | |||||||||||
\(\ds 121 \, 174 \, 811 + 5 \times 30\) | \(=\) | \(\ds 121 \, 174 \, 961\) | which is the $6 \, 904 \, 742$nd prime |
But note that $121 \, 174 \, 811 + 6 \times 30 = 121 \, 174 \, 991 = 7^2 \times 2 \, 472 \, 959$ and so is not prime.
Inspection of tables of primes (or a computer search) will reveal that no longer such sequences are known.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $121,174,811$