[seqfan] Re: a(n)th term is twice a(n)
Jack Brennen
jfb at brennen.net
Tue Jul 28 20:23:05 CEST 2009
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/
>
>
More information about the SeqFan
mailing list