Re: have you seen this sequence?
From: B. Cloitre (abcloitre_at_wanadoo.fr)
Date: 11/02/04
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Date: 2 Nov 04 18:58:36 -0500 (EST)
[snip]
"Experimentally it seems that the supremum of the n first terms
of the sequence grows like sqrt(4n/3)."
Seems true. Let (a(n))_{n>=1} denotes the considered sequence and let
M(n) be the supremum sequence :
M(n)=Max{a(k):1<=k<=n}
Then I believe there is this exact rule:
M(3k^2)=2k for k>=1
Max{a(k):1<=k<=3}=2
Max{a(k):1<=k<=12}=4
Max{a(k):1<=k<=27}=6
Max{a(k):1<=k<=48}=8
...
Does this inequality holds ?
abs(M(n)-sqrt(4n/3))<1
Regarding the density of 1's :
letting b(n)=#{1<=k<=n : a(k)=1} I observed :
0<=M(n)-b(n)<=1
I was wrong regarding the sequence of indices of occurrences of 1's in
(a(n)) : it is not the Favius sieve but it looks like.
This is truly an interesting sequence and I hope someone will like to
explore it further and prove above facts or other facts.
B. Cloitre
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