# [seqfan] Re: a(n), a(n+1), [a(n)+a(n+1)] share no digits

zak seidov zakseidov at yahoo.com
Mon Aug 31 19:32:40 CEST 2009

```Eric,

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 42, 51, 249, 306, 419, 2587, 3413, 5255, 6064, 7255, 8844, 11156, 22222, 31778, 60444, 72755, 88344, 111656,
222088, 333361, 422224, 508887, 622224, 708887, 922224, 5138383
base fini full

--- On Mon, 8/31/09, Eric Angelini <Eric.Angelini at kntv.be> wrote:

> From: Eric Angelini <Eric.Angelini at kntv.be>
> Subject: [seqfan]  a(n), a(n+1), [a(n)+a(n+1)] share no digits
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Date: Monday, August 31, 2009, 12:22 PM
>
> Hello SeqFans,
> we consider two consecutive integers ("a" and "b") from a
> monotonically increasing seq S, and their sum "c"; we want
> that:
>
> "a" shares no digit with "b"
> "b" shares no digit with "c"
> "c" shares no digit with "a"
>
> Starting with 1, do we have:
>
> S = 1,2,3,4,5,6,7,8,9,11,22,33,42,51,249,302,449,... ?
> (to prolong S, take the smallest integer not leading to a
>
> Does S stop at some point?
> (is this old hat, BTW?)
> Best,
> É.
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

```