[seqfan] Re: Very bounded sequences

[israel at math.ubc.ca]

P(n) = a(0)...a(n-1) grows exponentially, so heuristically the probability 
that 1+2*P(n-1) and 1+3*P(n-1) are both prime (resulting in a(n+1)>=4) is 
on the order of n^(-2). Since sum_n n^(-2) converges, we should expect this 
to occur only finitely many times. Of course this is not a proof. But if 
the reason your conjecture is true is "only statistical", maybe we should 
not expect a proof to be possible.

Cheers,
Robert

On Apr 10 2019, Tomasz Ordowski wrote:

>Dear SeqFans,
>
>I defined an interesting sequence:
>
>a(0) = 3; a(n) = smallest k > 1 such that 1 + a(0)a(1)...a(n-1)k is
>composite.
>
>3, 3, 3, 2, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2,
>3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, ...
>
>a(n) = 4 for n = 4 and 39,
>a(n) = 3 for n = 0, 1, 2, 5, 6, 7, 14, 20, 25, 56, 90, 119, 316, 330, 1268,
>1604, 1805, 1880, 1984, 2950, 3386, 3712, 4532, 4874, 8968, 18178, 19454,
>...
>a(n) = 2 for others n < 20000.
>Data from Amiram Eldar.
>
>Similar tails have sequences with other initial terms that are natural
>numbers.
>
>Conjecture: For any initial term a(0) > 0, a(n) > 3 only for finitely many
>n >= 0.
>
>The question is how to prove that all these sequences are bounded, so
>bounded?
>
>It seems that a(0) = 21 is the smallest initial term such that a(n) = 2 or
>3 for every n > 0.
>
>Note that if a(0) is a Sierpinski number, then a(n) = 2 for every n > 0.
>
>How to explain the occurrence of only twos and threes in the tails?
>
>Are similar sequences described in the literature?
>
>Best regards,
>
>Thomas Ordowski
>
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>
>