[seqfan] Re: As much as I hate "base" sequences...

[David Wilson]

For a = A235601, let S(0) = {1}, then for each n >= 1, compute set S(n) of
possible predecessors of elements of S(n-1).  Then a(n) is the smallest
element of S(n). Using this approach, I was able to compute up to a(100),
and could have gone further, but the elements were getting very large. I
completed A235601 and added a b-file for a(0..100).

For my original idea, where the target value need not be 1 (listing below),
the only values I verified were a(1) through a(8) = 1008000000. I can
provide strong statistical evidence that indeed a(n) = 1008*10^(n-2) for n
>= 8, but no proof. I can't use the same method I used to compute A235601 to
compute elements of this sequence, since I can't bound the size of the
targets, hence I can't initialize S(0).

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Neil
> Sloane
> Sent: Saturday, January 18, 2014 12:50 PM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: As much as I hate "base" sequences...
> 
> David, After looking more closely at your sequence, it seems that there is
> another version which is also nice: restrict the search to numbers which
> eventually reach 1.
> Of those, take the smallest that takes n steps to reach 1. That is a(n).
> The start is the same as your original version, 1, 2, 12, 108, 1944, 52488
but it is
> a different sequence, because 1008000000 never reaches 1.
> I added this version as A235601 under both our names - hope that is OK.
> Can you compute some more terms?
> Best regards
> Neil
> 
> 
> On Fri, Jan 17, 2014 at 7:26 PM, David Wilson
> <davidwwilson at comcast.net>wrote:
> 
> > Start with k and repeatedly apply the function
> >
> > k -> k / sum of digits of k
> >
> > stopping when there is a positive remainder or the divisor is 1.
> >
> > The smallest survivors of n iterations among the 29-smooth numbers are
> >
> > 0 1
> > 1 2
> > 2 12
> > 3 108
> > 4 1944
> > 5 52488
> > 6 1102248
> > 7 44641044
> > 8 1008000000
> > 9 10080000000
> > 10 100800000000
> > 11 1008000000000
> > 12 10080000000000
> > 13 100800000000000
> > 14 1008000000000000
> > 15 10080000000000000
> > 16 100800000000000000
> > 17 1008000000000000000
> > 18 10080000000000000000
> >
> > I am all but certain that these are these are indeed the smallest
> > survivors among the integers, and that the sequence extends to
> > infinity in the obvious way.
> >
> > The change in behavior at a(8) surprised me at first. a(1) through
> > a(7) eventually reach 1.  For n >= 8, we have
> >
> > a(n) = 1008*10^(n-2) ->  112*10^(n-2) -> 28*10^(n-2) -> 28*10^(n-3) ->
...
> > -> 28.
> >
> > ending at 28 after n iterations.
> >
> >
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
> 
> 
> 
> --
> Dear Friends, I have now retired from AT&T. New coordinates:
> 
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
> 
> _______________________________________________
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