This sequence is infinite but still does not contain all the primes. There is no way for 5 to appear, nor any higher prime starting with 5. -
Alonso del Arte, Sep 19 2015
Moreover, it is an obvious fact that there is no way for any prime starting with 2 (aside from the first two), 4, 6 or 8 to appear. -
Altug Alkan, Sep 20 2015
Apart from the first two terms, this sequence is identical to how it would be if it were to start with 5 and 53 instead of 2 and 23. -
Maghraoui Abdelkader, Sep 22 2015
We can initiate with a different prime p (see the a-file):
p=3: [a(3), a(4), ...];
p=5: [5, 53, a(3), a(4), ...];
p=7: [7, 71, 11, 13, 3, 31, 17, 73, 37, ...];
etc.
Define p~q to mean that the sequences generated by p and q eventually coincide (with different offset allowed). For example, we can see that 2~3~5, but it appears these are not equivalent to 7. Empirically, there are exactly four equivalence classes of primes:
Starting with 1, or starting with 2/4/5/6/8 and ending with 1
[11, 13, 17, 19, 41, 61, 101, 103, 107, 109, 113, 127, 131, 137, ...];
Starting with 3, or starting with 2/4/5/6/8 and ending with 2/3/5
[2, 3, 5, 23, 31, 37, 43, 53, 83, 223, 233, 263, 283, 293, 307, ...];
Starting with 7, or starting with 2/4/5/6/8 and ending with 7
[7, 47, 67, 71, 73, 79, 227, 257, 277, 457, 467, 487, 547, 557, ...];
Starting with 9, or starting with 2/4/5/6/8 and ending with 9
[29, 59, 89, 97, 229, 239, 269, 409, 419, 439, 449, 479, 499, ...].
(End)