# Sum three terms & digits

Eric Angelini keynews.tv at skynet.be
Thu May 24 11:50:37 CEST 2007

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Hello Max, Jonathan and Peter

I think Peter has computed exactly what I was looking for -- thanks a lot
to him. I am very sorry not being able to express in good english mathema-
tical words and formulas what I intend, sometimes.

My question is still open: do you believe this seq. is finite?

And what about sequences with the same constraint on _two_ consecutive
terms ?

Or four? Or five?

If the two-term rule, the three-term rule, the four-term rule don't lead
to finiteness, is there a n-term rule that does? For what smallest value
of n?

The starting seeds are relevant, of course: for the two-term addition rule
the seeds 1 and 203456789 block immediately the sequence ;-!

S = 1, 2034567890, ... END

Best,
É.

-----Message d'origine-----
De : Peter Pein [mailto:petsie at dordos.net]
Envoyé : mercredi 23 mai 2007 23:40
À : Max Alekseyev
Cc : Jonathan Post; Eric Angelini; seqfan at ext.jussieu.fr
Objet : Re: Sum three terms & digits

Max Alekseyev schrieb:
> That seems to be some corruption in the message text.
> E.g.: "15434339" in my original message was two numbers separated by a
> space:
> "1543" and "4339".
>
> Below I have added commas to separate terms of the sequence and
> included two more terms:
>
> 1, 2, 3, 4, 5, 7, 8, 9, 10, 14, 15, 31, 32, 33, 34, 35, 37, 38, 39,
> 40, 143, 339, 340, 1543, 4339, 14118, 31613, 31819, 136568, 231613,
> 331819, 3511268, 3556913, 32931819, 33511268, 333556913, 333631819,
> 1333511315, 2332856866, 3333631819, 35111211315, 35555156866,
> 329333631819, 335111211315, 3335555156866, 3336333631819,
> 13335111411315, 53328554956866, 633336333631819, 6313335111411315,
> 63123328554956866, 631263336333631819, 6312613335111411315,
> 63126123328554956866, 631261263336333631819, 6312612613335111411315,
> 63126126123328554956866, 631261261263336333631819,
> 6312612612613335111411315, 63126126126123328554956866,
> 631261261261263336333631819, 6312612612612613335111411315
>
> Max
>
> On 5/23/07, Jonathan Post <jvospost3 at gmail.com> wrote:
>> You wrote:
>>
>> ... 340 15434339 14118...
>>
>> and later:
>>
>> ... 3293181933511268  333556913...
>>
>
>

Hi Max,

I think you misunderstood Eric's intention (or I did so). He wrote:
"9+10+14=33 and "33" shares nothing...". Therefore the union of the
lhs-digits
shall have an empty intersection with the rhs-digits (10 and 14 would share
the "1").

My Mathematica-approach is:

noCommonDigit[{a_, b_}] :=
Module[{c = b + 1, abDigits = IntegerDigits[{a, b}]},
While[
Intersection[Flatten[{abDigits, IntegerDigits[c]}],
IntegerDigits[a + b + c]] =!= {},
c++];
c]

Prepend[
Last /@ NestList[Through[{Last, noCommonDigit}[#1]] & , {1, 2}, 100],
1]

{1, 2, 3, 4, 5, 7, 8, 9, 10, 14, 15, 31, 32, 33, 34, 35, 37, 38, 39, 40, 43,
44, 63, 64, 65, 68, 69, 73, 76, 79, 80, 83, 86, 88, 96, 116, 118, 119, 120,
124, 125, 128, 140, 267, 426, 440, 445, 446, 447, 460, 474, 604, 733, 774,
775, 777, 778, 779, 785, 797, 818, 819, 830, 873, 888, 889, 890, 893, 894,
913, 915, 916, 939, 945, 977, 1111, 1114, 1128, 1148, 1224, 1227, 1229,
1400,
2704, 2729, 2732, 2939, 2940, 2972, 2973, 4223, 4320, 6608, 6623, 6680,
6688,
7743, 7760, 9608, 14854, 48310, 159058}

Regards,
Peter

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