[seqfan] Re: a(n)th term is twice a(n)

[Eric Angelini]

 [Jack Brennen]:
 > It appears that the sequence repeats the pattern (...)

 ... yes, I've suspected smthg like this after working on
seq like:

    Each term of S "says":
    --> The a(n)th term of S is a(n)+k
    To build S, always take the smallest available integer not 
    leading to a contradiction (and no repeats)

For k=1 or 2, or 3, or 4, ... patterns appear almost im-
mediately -- thanks Jack!

Best,
É.





-----Message d'origine-----
De : seqfan-bounces at list.seqfan.eu 
[mailto:seqfan-bounces at list.seqfan.eu] 
De la part de Jack Brennen
Envoyé : mardi 28 juillet 2009 20:23
À : Sequence Fanatics Discussion list
Objet : [seqfan] Re: a(n)th term is twice a(n)

I assume that you're not allowing repeats in the sequence...

It appears that the sequence repeats the pattern:

a(6x) = 6x+1
a(6x+1) = 12x+2
a(6x+2) = 12x+4
a(6x+3) = 6x+5
a(6x+4) = 12x+8
a(6x+5) = 12x+10

with only a sparse set of exceptions...

All but three of the exceptions fit this pattern:

a(6*2^n) = 12*2^n
a(6*2^n+1) = 6*2^n+2
a(9*2^n) = 18*2^n
a(9*2^n+1) = 9*2^n+2

And three small exceptions seem to complete the sequence:

a(3) = 1
a(5) = 6
a(7) = 9

Eric Angelini wrote:
> Hello SeqFans,
> could someone please check and compute a few more terms of S?
> Each term of S "says":
> 
> --> The a(n)th term of S is 2*a(n)
> 
> To build S, always take the smallest available integer not 
> leading to a contradiction:
> 
> n = 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 ...
> S =  2  4  1  8  6 12  9 16 18 11 22 24 14 28 17 32 34 36 20 40 23 44 46 48 26 52 29 56 58 31 62 64 ...
> 
> Best,
> É.
> 
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 
> 



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