# Sum three terms & digits

Max Alekseyev maxale at gmail.com
Wed May 23 19:51:12 CEST 2007

```Eric,
The last listed element 42 should not be in this sequence since:39+40+42=121 share "2" with 42.
I have computed the following terms of this sequence so far:
1 2 3 4 5 7 8 9 10 14 15 31 32 33 34 35 37 38 39 40 143 339 340 15434339 14118 31613 31819 136568 231613 331819 3511268 3556913 3293181933511268 333556913 333631819 1333511315 2332856866 333363181935111211315 35555156866 329333631819 335111211315 33355551568663336333631819 13335111411315 53328554956866 6333363336318196313335111411315 63123328554956866 6312633363336318196312613335111411315 63126123328554956866 6312612633363336318196312612613335111411315 63126126123328554956866631261261263336333631819 631261261261333511141131563126126126123328554956866
Max
On 5/23/07, Eric Angelini <Eric.Angelini at kntv.be> wrote:>> Hello Alex,>> Yes, you are right:>> - seq S is monotonically increasing> - S starts with 1,2,3 (or 0,1,2)> - the next term you want to add to S must be the smallest>   one not infringing the rule>> I think this defines better S.>> Best,> É.>>>> -----Message d'origine-----> De : Max Alekseyev> [mailto:maxale at gmail.com]> Envoyé : mercredi 23 mai 2007 14:44> À : Eric Angelini> Cc : seqfan at ext.jussieu.fr> Objet : Re: Sum three terms & digits>> Eric,>> It is not clear how the sequence is defined.> Say, why it starts as 1,2,3,... and not as 1,2,1,... or as 1,2,2,... ?> Can you give a precise definition of this sequence?>> Max>> On 5/23/07, Eric Angelini <Eric.Angelini at kntv.be> wrote:> >> >> > Hello Seq-Fans,> >> > Is this sequence finite? I guess yes:> >> > S = 1,2,3,4,5,7,8,9,10,14,15,31,32,33,34,35,37,38,39,40,42...> >> > Sum three consecutive terms of S: the result cannot share> > any single digit with any of the three considered terms.> >> > Examples:> >> > 1+2+3=6 and "6" has no common digit with 1 or 2 or 3> > 8+9+10=27 and "27" has no common digit with 8 or 9 or 10> >> > But:> >> > 9+10+11=30 and "30" shares the digit "0" with "10" -- thus> >            "11" is not written after 10> > 9+10+12=31 and "31" shares the digit "1" with "10" (and "12")> >            thus "12" is not written after 10> > 9+10+13=32 and "32" shares "3"... thus... no "13"> >> > Now:> >> > 9+10+14=33 and "33" shares nothing, thus 33 is written after> >            9,10, etc.> >> > Best,> > É.> >> >> >>>

A024352 matches what I was going for with sequence two, and my first
well, I'll recompute and check (if not, I'll submit it). I don't want to do
all of this duplicate hunting and find out that I've submitted duplicate

-----Original Message-----

In general, it is easy to see that x^2 = y^2 + m has no integer
Therefore, the first one of your sequences is simply A016825:

%C A016825 No solutions to a(n)=b^2-c^2 - Henry Bottomley
(se16(AT)btinternet.com), Jan 13 2001

The second sequence is A042965 or A024352, whichever you like.

Max

On 5/23/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:
> Good point,
>
> 3 = 1(1+2)
> 5 = 1(1+4)
> 7 = 1(1+6)
> etc.
>
> So all odd numbers should, trivially, be in the sequence. Thanks!
>
>         -Andrew Plewe-

```