[seqfan] Re: Correct seq defined by: a(a(n)) is a square
Benoît Jubin
benoit.jubin at gmail.com
Sun Nov 22 17:37:37 CET 2009
This sequence is characterized by:
* a(n) is either n+1 or a square
* all the squares appear and they appear in increasing order
* every other term is a square, except when the index is a square, in
which case, the term is also a square.
Alternatively, we can say that a(n)=n+1 exactly when n=3 or (n minus
the largest square at most n) = n - (floor(sqrt(n)))^2 is odd, else it
is the least square not already in the sequence.
This sequence can begin at 0 since no special treatment is needed.
Benoit
On Sun, Nov 22, 2009 at 4:56 PM, Eric Angelini <Eric.Angelini at kntv.be> wrote:
>
> Hello SeqFans (quater)
> Here is the correct seq obeing to: "a(a(n)) is a square":
> S = 1,3,4,9,6,16,8,25,36,11,49,13,64,15,81,100,18,121,20,144,22,169,24,196,225,27,256,29,289,31,324,33,361,...
> Again, "always use the smallest available
> integer not leading to a contradiction".
> I guess that more of half the terms are
> squares, and that this seq is "densier"
> (in term of squares) than this one (see
> former post): a(a(n))=a(n) squared
>
> ...
>
> Best,
> E.
>
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