[seqfan] Re: Interesting sequence

Andrew Weimholt andrew.weimholt at gmail.com
Sun Mar 6 22:56:44 CET 2016


On Sun, Mar 6, 2016 at 1:10 PM, David Wilson <davidwwilson at comcast.net>
wrote:

> a(1) = 1
> a(n+1) =
>     [sqrt(n)] if not already in sequence
>     [sqrt(n)] + n otherwise.
>
> Is this sequence a permutation of the positive integers?
>
>
Yes, it is a permutation of the positive integers.
1) If k hasn't yet occurred in the sequence by the time we get to a(k^2),
then it will occur at a(k^2+1).
2) We'll never be forced to include a repeat, because for any k,m with k<m,
[sqrt(k)] < [sqrt(k)]+k < [sqrt(m)]+m,
therefore a(k) < [sqrt(m)]+m, so if [sqrt(m)] is already taken then a(m)
will be greater than any previous value.

This holds for [x] = floor(x) as well as [x] = ceiling(x).
So we actually have 2 sequences from this idea (4 if you count the inverse
permutations).

Andrew


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