[seqfan] Re: Exponentially S-numbers (set of all densities)

[Vladimir Shevelev]

Dear Seq Fans,

Let S be a finite or infinite increasing sequence
of integers beginning  from 1;  E(S) be the 
sequence of integers having all exponents in 
prime power factorization from S; h(E(S)) be 
the density of E(S).  In
http://arxiv.org/pdf/1510.05914v3.pdf
I proved that h(E(S)) always exists and found
an expression for it. It is clear that h(E(S))
is in interval [1/zeta(2), 1]. As I already
wrote, answering on my question, if the
set {h(E(S))} is a dense set in this interval,
D. Berend found a gap in {h(E(S))}.
But the structure of {h(E(S))} remained
open. Now in
http://arxiv.org/abs/1511.03860v1.pdf
I proved that {h(E(S))} is a perfect set
with a countable set of gaps which
associate with some left-sided neighborhoods
of the densities of all exponentially finite S-
sequences, except for S={1}.
It is yet unknown, if the sum of 
lengths of all gaps equals the length of the
whole interval [1/zeta(2), 1], or, the
same, if the set {h(E(S))} has zero
measure?
 Note that the Cantor perfect set was 'man-made',
while this set arised naturally. In this case I did 
not even think that this set can have such a 
complicated and interesting structure.

Best regards,
Vladimir


________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 27 October 2015 13:49
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Exponentially S-numbers (set of all densities)

Dear Seq Fans,

Recall that I found the density of the exponentially
S-numbers  for every increasing sequence S of positive
integers. It is evident that, for all such sequences
beginning from 1, the densities of the exponentially
S-numbers are in the interval [1/zeta(2), 1]. In my
paper I asked, whether the set of all such densities
is a dense set in this interval? Daniel Berend (private
communication) gave a negative answer. He found
a gap in the set of all densities of length approx. 0.083.
It is a large gap, taking into account that the length
of the whole interval [1/zeta(2), 1] is approx. 0.39.
But more important another thing. Using his idea,
we can find infinite chains of gaps. It would be
very interesting to find the sum of all lengths of the
gaps. In particular, it would be nice if it equals
1-1/zeta(2). Then our set would be similar to Cantor
set. I placed Berend's proof in my paper (Section 6)
which is available at

http://arxiv.org/abs/1510.05914


Best regards,
Vladimir



________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 21 October 2015 16:27
To: seqfan at list.seqfan.eu
Subject: [seqfan] Exponentially S-numbers

Dear Seq Fans,

When I learned about A209061 (exponentially squarefree
numbers, or the numbers having all squarefree exponents
in their prime power factorization), I paid attention on a
remarkable formula in Toth's link (Theorem 3, 2007): if RH is true,
then
Sum_{a(n)<=x} 1= hx+O(x^(0.2+eps)),  (1)
for every eps>0,
h=Prod{p}(1+Sum_{k>=4} (mu^2(k)-mu^2(k-1))/p^k) (2),
where product is over all primes, mu is the Möbius function.
Later I found many other papers, beginning with 1972, but
in all of them the authors considered only exponentially squarefree
numbers and, using or not RH, tried improve the remainder term.
Indeed, a deep M. V. Subbarao's 1972-paper,
where he gave some general constructions and for a first time
intoduced the notion " exponentially squarefree
numbers" made this topic classical.  Formula (1) has practically
the best known remaider term.
I had a dream: to find a generalization of (1)-(2), maybe with
less good remainder term, but is suitable for the
exponentialy S-numbers for every fixed increasing sequence S
of positive integers. My approach is quite another, than all these
authors. I based on my paper "Compact numbers and factorials",
Acta Arith. 126(2007), no.3 , 195-236, Theorem 1.
I understood that "compact numbers" one can also
name exponentially 2^n-numbers. This was the first clue.
Thank God, I was able to generalize this very
special case. I obtained a common for exponentially
S-numbers (i.e., independent on S) remainder term equals
O(sqrt(x)*log(x)*e^(c*sqrt(log(x))/log(log(x)))), where
c=4*sqrt(2.4/log(2))=7.4430...,
i.e., O(x^((1/2)+eps) with concretizing eps.

Now I am pleased to inform Seq Fans that to-day my paper
with a general case has been appeared in arxiv :

http://arxiv.org/pdf/1510.05914v1.pdf

Best regards,
Vladimir







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