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

[franktaw at netscape.net]

The next term can have at most one digit more, so 100 is an upper bound.

Franklin T. Adams-Watters


-----Original Message-----
From: Jack Brennen <jfb at brennen.net>

Brings up some interesting side questions, such as whether there is a
bound on the ratio a(n+1)/a(n).  (Assume that a(n) can be arbitrary;
it doesn't have to be reached from a smaller number.)

For instance, if a(n) is 13427, a(n+1) is 886668, for a ratio of
just over 66.  How high could this ratio get?

   Jack

zak seidov wrote:
> Eric,
> your sequence terminates at a(38)=5138383:
>
> 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
>>  contradiction)
>>
>> Does S stop at some point?
>> (is this old hat, BTW?)
>> Best,
>> É.