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

[Eric Angelini]

Hi Bob,
My post was sent while I was running 
to the movie theatre, sorry!
"Onwards" we end the "infinite" 3-digit figures with:
..., 835,843,851,859,867,875,883,891,900 (and _not_ 899, which is also possible), 909,918,927,936,945,954,963,972,981,990,999. The next integer is 1000.

If you check, you'll see that:
900+9=909,
909+9=918,
918+9=927, etc. -- the single added '9' being the leftmost digit of the result.
But Jack closed the case!
Best,
É.
Catapulté de mon aPhone

> Le 30 août 2014 à 21:51, "Bob Selcoe" <rselcoe at entouchonline.net> a écrit :
> 
> 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
>> 
>> _______________________________________________
>> 
>> Seqfan Mailing list - http://list.seqfan.eu/
> 
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