919 conjecture

[Tanya Khovanova]

On 7/17/07, Leroy Quet <qq-quet at mindspring.com> wrote:

> Can it be PROVED that all terms of A131790 are finite, possibly using
> Hardy and Wright or some other such reference?

It would follow from the statement:

There exist infinitely many positive integers k such that both 2k+1
and 3k+1 are prime.
I do not have a proof that this statement is true (it may be as hard
as twin prime conjecture) but heuristic arguments ala Hardy--Wright
suggest that it is ture.

Now, if for some k>1 both 2k+1 and 3k+1 are prime then 3(2k+1)=6k+3
and 2(3k+1)=6k+2 both have exactly 4 divisors, implying that
A000005(6k+3)=A000005(6k+2)=4 belong to the same run in A000005 and
the length of this run is greater than 1. In other words, in A131789
elements greater than 1 appear infinitely often.
It is also clear that for each prime p >= 5, A000005(p)=2 forms its
own run of length 1 in A000005, implying that in A131789 elements
equal 1 appear infinitely often.
Therefore, all runs in A131789 are finite and so are elements of A131790.

P.S. btw, the sequence of k such that both 2k+1 and 3k+1 are prime
seems to be missing in OEIS. It starts with:
2, 6, 14, 20, 26, 36, 50, 54, 74, 90, 116, 140, 146, 174, 200, 204,
210, 224, 230, 270, 284, 306, 330, 336, 350, 354, 384, 404, 410, 426,
440, 476, 510, 516, 554, 564, 596, 600, 624, 644, 650, 704, 714, 726,
740, 746, 834, 846, 894, 930, 944, 950

Regards,
Max