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

[Neil Sloane]

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.
>
>
>
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>
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>



-- 
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.
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