[seqfan] Re: Onwards a(n) = a(n-1) + first digit of a(n)

[Jack Brennen]

Why is it not possible to compute it infinitely without backtracking?  Seems like each time
that you cross to a power of 10^N (N>=3), you have to choose one of six paths to take to the
next one, the right path depending on the value of N modulo 6, and designed to let you hit
8*10^N-1.

 From 10^N, you can choose to hit either 2*10^N or 2*10^N+1.
 From those, you can choose to hit anything from 3*10^N to 3*10^N+2.
 From those, you can choose to hit anything from 4*10^N to 4*10^N+3.
 From those, you can choose to hit anything from 5*10^N to 5*10^N+4.
 From those, you can choose to hit anything from 6*10^N to 6*10^N+5.
 From those, you can choose to hit anything from 7*10^N+1 to 7*10^N+6.
 From those, you choose the right one to allow you to hit 8*10^N-1.
 From 8*10^N-1, you can hit 9*10^N-9.
 From 9*10^N-9, you can hit 10*10^N-1.

But the path from 10^N to the right 7*10^N+X is predetermined for you by the
value of N modulo 6.

Am I missing something subtle?

   Jack


On 8/30/2014 9:44 AM, Eric Angelini wrote:
>
> Hello SeqFans,
> S=0,9,10,11,12,13,14,15,16,17,18,20,22,24,26,28,31,34,37,41,45,49,54,60,66,73,81,90,99,100,101,102,...
>
> S was computed backwards, starting with a(1)=1000 and
> always subtracting to a(n) it's first digit.
> Thus 1000,999,990,891,883,875,867,..., 12,11,10,9,0.
>
> Is it possible to compute an infinite S onwards, starting with 0, without backtracking? No!
> This is fascinating...
>
> Best,
> É.
>
> Catapulté de mon aPhone
>
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