[seqfan] Re: Simple seq with 3 in 4
Alex M
timeroot.alex at gmail.com
Sat Aug 21 19:20:02 CEST 2010
Simplest formula I could get: a(3n)=n-1; a(3n+1)=n+1; a(3n+2)=n+3. Or,
a(n)=1+a(n-3).
~6 out of 5 statisticians say that the number of statistics that either make
no sense or use ridiculous timescales at all has dropped over 164% in the
last 5.62474396842 years.
On Sat, Aug 21, 2010 at 9:30 AM, Eric Angelini <Eric.Angelini at kntv.be>wrote:
> Hello SeqFans,
> I must have missed the correct entry
> in the OEIS because I cannot find this
> easy (core?) seq.
> "To find 3 consecutive naturals in S,
> you have to take 4 consecutive terms
> of S -- no less":
> S= 1,3,0,2,4,1,3,5,2,4,6,3,5,7,4,6,8,5,7,9,6,8,10,7,9,11,8,10,...
>
> Ex: taking the first 3 terms doesn't
> allow you to handle 3 consecutive
> natural numbers as they are 0,1 and... 3.
> But if you take the fourth term (2),
> you'll have in hand 0,1,2 [and even
> another triplet of consecutive naturals,
> which is (1,2,3)].
>
> Formula is easy to compute.
>
> Best,
> E.
>
>
>
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