[seqfan] Re: Simple seq with 3 in 4

[Alexander P-sky]

PURRS Demo Results
Verified exact solution for x(n) = 1+x(-3+n)
for the initial conditions
x(0) = 1
x(1) = 3
x(2) = 0
Verified solution x(n) =
1+1/3*n-(1/18*I)*sqrt(3)*(-1/2+(1/2*I)*sqrt(3))^n+1/2*(-1/2-(1/2*I)*sqrt(3))^n+1
/2*(-1/2+(1/2*I)*sqrt(3))^n+(1/18*I)*sqrt(3)*(-1/2-(1/2*I)*sqrt(3))^n+(-1/2-(1/2
*I)*sqrt(3))^(-1)*(-1/2-(1/2*I)*sqrt(3))^n+(-1/2+(1/2*I)*sqrt(3))^(-1)*(-1/2+(1
/2*I)*sqrt(3))^n
for each n >= 0

On 8/21/10, Alex M <timeroot.alex at gmail.com> wrote:
> 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.
>>
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>