[seqfan] Re: Dividing a concatenation

[israel at math.ubc.ca]

It's not a permutation of the positive integers, because for n > 1 no term 
can be divisible by 2 or 5.

Cheers,
Robert

On May 4 2015, Eric Angelini wrote:

>Hello SeqFans,
>we want a(n) to be the smallest unused integer in S so far;
>we want a(n) to divide the concatenation of [a(1),a(2)... a(n-1)]
>
>S = 2,1,3,71,7,10177,2100001,101770000001,4603,13,107,4013,23,...
>
>Explanations:
>
>S cannot start with "1" -- thus a(1)=2;
>
>2            (start)
>1            is the smallest integer (not yet present in S) dividing 2 
>3            is the smallest integer (not yet present in S) dividing 21
>71           is the smallest integer (not yet present in S) dividing 213
>7            is ... dividing 21371
>10177        is ... dividing 213717
>2100001      is ... dividing 21371710177
>101770000001 is ... dividing 213717101772100001
>4603         is ... dividing 213717101772100001101770000001
>13           is ... dividing 2137171017721000011017700000014603
>107          is ... dividing 213717101772100001101770000001460313
>4013         is ... dividing 213717101772100001101770000001460313107
>23           is ... dividing 2137171017721000011017700000014603131074013
>
>I guess the next terms are (not sure):
>
> 3097349301044927552199565217412468305904367 1847 37 
> 57761378857324324622100000000394679218216573867495927055268312221609870740877521781199113101 
> 73 9 27 243 11 201010661 19 49 21 ...
>
>Is S a permutation of the positive integers?
>
>Best,
>É.
>
>
>
>
>
>
>
>
>
>_______________________________________________
>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>