# [seqfan] Re: |a-b| divides concatenation [ab]

Jack Brennen jfb at brennen.net
Mon Sep 20 19:40:43 CEST 2010

```Also note that brute force runs into some issues around term 446
and beyond.  You can brute force the first 445 terms of T without
too much trouble, but the 446th term is 4545970,
and the 451st term is 142857750.

Once brute force runs out of steam, it becomes an exercise in
looking for factors of T[n]*(10^y+1) for small y.  For instance,
the 450th term is 607, and the 451st term is found to be
142857750 because 142857143 is a divisor of 607*(10^9+1), and
thus 142857143+607 = 142857750 works to continue the sequence.

Using the factoring method is probably the best way to compute
the sequence; it'll be a little slower than brute force for the
early terms, but probably not terribly so, and it'll be orders
of magnitude faster once you get into the bigger numbers.

Jack Brennen wrote:
> Eric Angelini wrote:
>> If we drop the "strictly increasing" constraint, we'll get T (which
>> is an incredible nightmare to calculate by hand):
>>
>>      T = 1, 144, 43, 134, 108, 9, 6, 4, 158, ...
>> 1st dif:  143 101  91   26  99  3  2  154 ...
>>
>
> Not easy by hand, I guess, because I think you're off on the fourth
> term, and everything past that is bogus then.  :)
>
> I'm assuming that the sequence T and the difference sequence are
> supposed to be disjoint, and that each subsequent term of T is the
> smallest possible value that preserves this property?  Assuming that,
> I get (with computer assistance):
>
> 1   144   43  120  80  60  5   390   87  58 56  42  9 6   160
>   143   101  77   40  20  55 385   303  29  2  14  33 3 154
>
> I think I have 450 terms of T, but I'll hold off on posting them
> until I get confirmation that the sequence looks right through
> those first 15 terms.
>
>
>
>
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>

```