[seqfan] Re: Correct seq defined by: a(a(n)) is a square
benoit.jubin at gmail.com
Sun Nov 22 17:48:12 CET 2009
and when a(n) is a square, it is the square of (n + floor(sqrt(n)))/2.
(sorry for the double email)
Summing up: a(n)=n+1 if n-(floor(sqrt(n)))^2 is odd and
[(n+floor(sqrt(n)))/2]^2 if n-(floor(sqrt(n)))^2 is even.
2009/11/22 Benoît Jubin <benoit.jubin at gmail.com>:
> 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.
> 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
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