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

[Bob Selcoe]

Hi Eric,

I just want to make sure I understand what you're saying.

I think you mean  1000, 999, 990... 909, 900, 891, 883, 875, 867,..., 
12,11,10,9,0.   Yes?

If so, then if a(1) = 1000, the equation is a(n) = a(n-1) - first digit of 
a(n-1)  ???

Also,  not sure what you mean by "backtracking".

Cheers,
Bob Selcoe


--------------------------------------------------
From: "Eric Angelini" <Eric.Angelini at kntv.be>
Sent: Saturday, August 30, 2014 11:44 AM
To: "Sequence Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Onwards a(n) = a(n-1) + first digit of a(n)

>
> 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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