[seqfan] Re: A monotonically increasing sequence with a(a(a(n)))being odd
Eric Angelini
Eric.Angelini at kntv.be
Mon Nov 30 18:48:18 CET 2009
> anticipating
... indeed, thanks, Benoît !
Me:
> I'm curious to see a monotonically increasing sequence with
> a(a(a(n))) being prime.
... this seems impossible if we want both the seq to be mono-
tonically increasing AND to show all primes
Best,
É.
-----Message d'origine-----
De : seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu] De la part de Benoît Jubin
Envoyé : lundi 30 novembre 2009 18:18
À : Sequence Fanatics Discussion list
Objet : [seqfan] Re: A monotonically increasing sequence with a(a(a(n)))being odd
Anticipating your next email, I computed the two sequences:
*smallest injective sequence such that a(a(...a(n))) (with n
applications of a) is even, with offset 0. This turns out to be
A014681. For offset 1, this is the same sequence truncated, A103889.
*smallest injective sequence such that a(a(...a(n))) (with n
applications of a) is odd, with offset 1 (since there is no such
sequence with offset 0). This gives 1,3,5,2,7,4... which is not in
the OEIS. It is equal to A065168 and A165754 from the third term on
(that is, a(2n)=2n-2 and a(2n-1)=2n+1 for n>=2).
Benoit
On Mon, Nov 30, 2009 at 4:35 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> Hello SeqFans,
>
> The seq starts :
>
> S = 1,3,4,5,7,8,9,10,11,13,15,16,17,18,19,20,21,22,23,25,27,29,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,49,51,53,55,57,59,61,63,64,65,66,...
>
> Looking at runs of consecutive integers we see that we have
> (after the first integer, "1"):
>
> 1st run length is 3 (= 2^1 + 1) 3,4,5,
> 2nd run length is 5 (= 2^2 + 1) 7,8,9,10,11,
> 3rd run length is 9 (= 2^3 + 1) 15,16,17,18,19,20,21,22,23,
> 4th run length is 17 (= 2^4 + 1) 31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,
> ... etc.
>
> Those runs are separated by runs of consecutive odd integers:
>
> 1st run length is 0 (= 2^0 - 1) -
> 2nd run length is 1 (= 2^1 - 1) 13
> 3rd run length is 3 (= 2^2 - 1) 25,27,29,
> 4th run length is 7 (= 2^3 - 1) 49,51,53,55,57,59,61,
> ... etc.
>
> So a general formula for a(a(a(n))) is not impossible to find.
>
> Missing terms in S are:
>
> M(s) = 2,6,12,14,24,26,28,30,48,50,52,54,56,58,60,62,...
>
> ---
>
> I'm curious to see a monotonically increasing sequence with
> a(a(a(n))) being prime.
>
> Best,
> É.
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
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