# [seqfan] Re: Numbers n such that (n+n+1) divides the concatenation [n, n+1]

Robert Israel israel at math.ubc.ca
Tue Sep 21 19:23:15 CEST 2010

```The sequence is 1, 4, 16, 49, 166, 499, 1666, 4999, ...

If n+1 has k digits, the concatenation is 10^k n + n + 1, which is
congruent to n (10^k - 1) mod 2n+1.  Since n and 2n+1 are coprime,
the condition is just that 2n+1 divides 10^k - 1.  Now n+1 having k
digits means 10^(k-1)  <= n+1 <= 10^k - 1, so d = (10^k-1)/(2n+1) can only
be 1, 2, 3, 4 or 5.  But it can't be 2, 4 or 5 because 10^k - 1 is
not divisible by 2 or 5.  The case d=1 corresponds to n= 5*10^(k-1) - 1
(thus 4, 49, 499, ...) and the case d=3 corresponds to n = (10^k-4)/6
(thus 1, 16, 166, 1666, ...).

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada

On Tue, 21 Sep 2010, Eric Angelini wrote:

>
> I think this seq (not in the OEIS) starts like this:
>
> C = 1,4,16,49,166,...
>
> 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?
> Best,
> É.
>
>
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>
```