Notes on A062316 et al

[David Wilson]

Concerning the following:

%F A062316 a(n) is always even. It seems that a(n+1)-a(n)=4 or 8 only. Does 
limit n-->infinity a(n)/n = 6 ? If so, does\ a(n)-6*n = 0(log(n)) ? - Benoit 
Cloitre (abcloitre(AT)wanadoo.fr), Feb 16 2003

A062316 is, by definition, the intersection of the numbers which are not 
sums of two squares (A022544) and the numbers which are not the difference 
of two squares (A016825). A016825 consists precisely those n == 2 (mod 4), 
so the above observation that a(n) is always even can be sharpened to a(n) 
== 2 (mod) 4.

The observation that the first difference is 4 or 8 appears to be correct. 
The boundedness of the first differences of A062316 would imply the 
boundedness of first differences of A022544, so I went back and checked, and 
it appears that the first differences of A022544 are always 1, 2, 3, or 4. I 
have no explanation, but I am sure it is true.  Maybe someone better versed 
in number theory could sort this out.

With regard to the conjecture about the limit of a(n)/n being 6, this is 
wrong. Empirically, I could see that as n increased, a(n)/n seemed to 
decrease slowly but fairly steadily, and I ran my program far enough out to 
convince myself that a(n)/n < 5.3 for sufficient n.

On further research, I found Mathworld's article "Landau-Ramanujan Constant" 
indicates that A001481 is a O(x/sqrt(ln(x))), so that A001481 has density 0 
over Z+. This means that its complement, A022544, has density 1, with limit 
a(n)/n = 1. This in turn means A016825 has density 1/4, with limit a(n)/n = 
4.

I would remove the above line from A062316 add the following lines to 
various sequences:

%C A001481 lim n->inf a(n)/n = inf.


%C A022544 lim n->inf a(n)/n = 1.
%C A022544 Conjecturally, a(n+1)-a(n) = 1, 2, 3 or 4.

%C A062316 Elements of A022544 congruent to 2 (mod 4); intersection of 
A022544 and A016825.
%C A062316 Conjecturally, a(n+1)-a(n) = 4 or 8.
%C A062316 lim n->inf a(n)/n = 4.



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