[seqfan] Re: Numbers n such that (n+n+1) divides the concatenation [n, n+1]
Alexander P-sky
apovolot at gmail.com
Tue Sep 21 19:14:49 CEST 2010
Input interpretation:
{1, 4, 16, 49, 166, 499, 1666, 4999, 16666, 49999, ...}
Possible closed form:
a_n = 1/12 (-2 (-1)^n+10^(n/2+1/2)-(-1)^n 10^(n/2+1/2)+3 10^(n/2)+3
(-1)^n 10^(n/2)-10)
(for all terms given)
On 9/21/10, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> On Tue, Sep 21, 2010 at 10:51 AM, Eric Angelini <Eric.Angelini at kntv.be>
> wrote:
>>
>> I think this seq (not in the OEIS) starts like this:
>>
>> C = 1,4,16,49,166,...
>>
>
> (10:46) gp > last=0;for(n=1,99999, eval(Str(n,n+1))%(1+2*n) |
> print1(n,","))
> 1,4,16,49,166,499,1666,4999,16666,49999,
>
>> We have:
>> 1+2 divides 12 -> 12/3=4
>> 4+5 divides 45 -> 45/9=5
>> 16+17 divides 1617 -> 1617/33=49
>> 49+50 divides 4950 -> 4950/99=50
>> 166+167 divides 166167 -> 166167/333=499
>> ...
>> Pattern?
>
> Yes! ;-)
> Proof : left as exercise.
>
> M.
> %F :
> a(2n)=5*10^(n-1)-1
> a(2n+1)=(10^n*5-2)/3
>
>
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