[seqfan] Re: Some large numbers

[Max Alekseyev]

This sequence continues to alternate between 2's and 3's, which
follows from the two statements:

For positive integers x,y:
(i) if x < 3^y, then 2^x < 3^(3^y).
(ii) if 2*y < 2^x, then 2*3^y < 2^(2^x)

The statement (i) is trivial, the statement (ii) is almost trivial:
2^(2^x) >= 2^(2*y+1) = 2*4^y > 2*3^y.

Now, we can start induction from 2^4 < 3^3 and 2*3^3 < 2^(2^4).

Regards,
Max

On Thu, Sep 27, 2012 at 8:26 AM,  <franktaw at netscape.net> wrote:
> Consider https://oeis.org/A014221 - a(n+1) = 2^a(n), and
> https://oeis.org/A014221 - a(n+1) = 3^a(n).
> Now merge the two sequences:
>
> 0, 1, 2, 3, 4, 16, 27, 65536, 7625597484987[, 2^65536]
>
> We're interested here in how the merge proceeds, so drop the initial 0, 1,
> and replace each subsequent term with it's base:
>
> 2, 3, 2, 2, 3, 2, 3[, 2]
>
> How does this sequence continue? Does it alternate from here between 2's and
> 3's, or are there more 2,2 substrings? Is there an efficient way to compute
> the sequence?
>
> Franklin T. Adams-Watters
>
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