[seqfan] Re: Interesting sequence
Neil Sloane
njasloane at gmail.com
Tue Nov 22 16:56:12 CET 2016
Back in 2004, Jeff Lagarias and I investigated what happens
when you start with a rational number x > 1, and repeatedly
apply the map x -> x times ceiling of x
See J. C. Lagarias and N. J. A. Sloane, "Approximate Squaring", *Experimental
Math.*, 13, Number 1 (2004), pp. 113-128; arXiv:math.NT/0309389
<http://arxiv.org/abs/math.NT/0309389>.
Available from http://neilsloane.com/doc/Me273.pdf
This is a different problem, of course, but kind-of related.
We conjecture that the process always reaches an integer. but this is still
an
open question. We noticed that numbers of the form (n+1)/n seemed to take
the most steps
before they reached an integer. (Which is why David's question reminded me
of this.)
For example starting with 200/199, the first integer that is reached is
200^(2^1444), which has about 10^435 digits.
Best regards
Neil
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
On Mon, Nov 21, 2016 at 10:14 PM, Robert G. Wilson v <rgwv at rgwv.com> wrote:
> Et al,
>
> Mmca: f[n_] := Block[{m = k = n}, While[k > 0, m = Floor[m (k + 1)/k];
> k--]; m]; Array[f, 100]
>
> RGWv
>
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of David
> Wilson
> Sent: Monday, November 21, 2016 8:40 PM
> To: 'Sequence Fanatics Discussion list'
> Subject: [seqfan] Interesting sequence
>
> Starting with integer n, multiply by ((n+1)/n), take the floor, multiply
> by (n/(n-1), take the floor, all the way down to 2/1, call the result f(n).
>
> For example, starting with n = 5
>
> floor(5*(6/5)) = 6,
> floor(6*(5/4)) = 7,
> floor(7*(4/3)) = 9,
> floor(9*(3/2)) = 13,
> floor(14*(2/1)) = 26.
>
> so f(5) = 26.
>
> Starting at n = 1, we have
>
> f = (2, 6, 12, 18, 26, 38, 48, 62, 78, 90, ...)
>
> It's trivial that all elements are even, given the final multiplier 2/1.
>
> It looks to me as if f(n) ~ pi*n^2/4, but I couldn't begin to prove this.
>
>
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