[seqfan] Re: Numbers with squarefree exponents in their prime factorizations
Vladimir Shevelev
shevelev at bgu.ac.il
Fri Sep 18 21:25:48 CEST 2015
Dear Seq Fans,
Sorry!! Peter has found
an error in his programs.
So, thank God, we have no
any "strange problem".
In particular, all representations
of Toth's constant give the same result!
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 15 September 2015 20:45
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Numbers with squarefree exponents in their prime factorizations
Dear Seq Fans,
We have here a strange problem which
formally could arise also in other areas
of mathematics. Let A(x) = b*x+ o(x)
(in our case, A(x) is the number of terms
of A209061 not exceeding x, o(x)=O(x^(0.2+eps)).
Suppose, we cannot calculate the constant b
better than up to accuracy 0.001. Then
we know A(x) with "true remainder term"
x/10000 and when o(x)<x/10000, it is
practically pointless.
We did with Peter Moses several experiments.
I asked him to calculate
Prod_{prime p}(1 - sum_{k>=2} (p-1)/p^(k+1)).
He gave answer 0.6029... It is not true, it should be
0.6079... Indeed, it is easy to prove that
sum_{k>=0} (p-1)/p^(k+1))=1,
so Prod_{prime p}(1 - sum_{k>=2} (p-1)/p^(k+1))=
Prod_{prime p}((p-1)/p + (p-1)/p^2)=Prod_{p}
(1 - 1/p^2)=1/zeta(2)=6/pi^2=0.6079.
In other case, I asked him to calculate
Prod_ {prime p}(sum_ {k is in S} (p-1)/p^(k+1)),
where S is the set of 0 and nonnegative powers of 2.
He gave the exact answer 0.872497... coinciding
with my for the density of so-called "compact numbers"
(in Theorem 1 in my 2007-paper "Compact numbers and
factorials" (Acta Arith, 126.3 (2007), 195-236, I used
quite another approach). Fatal errors, in my opinion,
arise when set S has large density, for example, S=0
and squarefree numbers. If S has zero
density (as here {2^n} or {n^2} or even {p_n}),
then I believe that we could obtain the nice results.
Since, as I showed, L. Toth's formula is the same one
that uses S of 0 and squarefree numbers, then I think
that it is difficult to calculate the density in A209061
or A130897 with the accuracy 0.0001. Indeed, it is better
to use the set with a smaller density. Therefore, it is better
to use the formula
Prod_ {prime p}(1 - sum_ {k is in S_1}((p-1)/p^(k+1)),
where S_1 is {not sqarefree numbers}, than
the formula
Prod_ {prime p}( sum_ {k is in S_2}((p-1)/p^(k+1)),
where S_2 is {0 and sqarefree numbers} (or the equivalent
Toth's formula).
These formulas should give the same result.
However, they give different results: by the first of them,
0.9556950..., by the second one, 0.95592301... .
Any thoughts are wellcome.
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 12 September 2015 23:22
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Numbers with squarefree exponents in their prime factorizations
Let us show that in fact we have the
same formula, i.e., (2) coincides with (1).
Consider Toth's p-factor:
1+sum_{j>=4}((mu(j)^2-mu(j-1)^2)/p^j ) =
1+sum_{j>=4}mu(j)^2/p^j -
1/p* sum_{j>=4}mu(j-1)^2)/p^(j-1)=
1+sum_{j>=5}mu(j)^2/p^j -
1/p* sum_{j>=3}mu(j)^2/p^j =
1-1/p^3 + sum_{j>=3}mu(j)^2/p^j -
1/p* sum_{j>=3}mu(j)^2/p^j =
1-1/p^3 + (1-1/p)*sum{j squarefree>=3}1/p^j=
1-1/p^3 + sum{j squarefree>=3}(p-1)/p^(j+1).
Let us compare this expression with my p-factor:
sum_{j runs 0 and squarefree numbers}
(p-1)/p^{j+1} = (p-1)/p +(p-1)/p^2 + (p-1)/p^3 +
sum{j squarefree>=3}(p-1)/p^(j+1).
But 1-p^3=(p-1)/p +(p-1)/p^2 + (p-1)/p^3.
So my factor coincides with Toth's one!
Since, beginning with the fourth digit, we have
the contradictory evaluations of the same product,
then there is a computing error.
Could anyone calculate the density
of exponentially squarefree nunbers
up to more than the third digit, i.e.,
better than 0.955?
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 11 September 2015 22:28
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Numbers with squarefree exponents in their prime factorizations
Dear Seq Fans,
For a fixed prime p, 1/p^j -1/p^(j+1) = (p-1)/p^(j+1)
is the density of the numbers divisible by p^j and not divisible
by p^(j+1), i.e., the numbers with p^j in their prime power
factorization (PPF). Let S be a set of nonegative integers.
Then sum_{j in S} (p-1)/p^(j+1) means the density of
the numbers having in their PPF the prime p with exponent
from S. Thus convergent product
prod_{prime p}(sum_{j in S} (p-1)/p^(j+1)) (1)
should give the density of the numbers with all exponents
from S in their PPF. For example, let S_k be set {0,1,...,k-1}.
Then sum_{j in S_k} (p-1)/p^(j+1) = 1 - 1/p^k and
prod_{prime p}(sum_{j in S} (p-1)/p^(j+1)) =
prod_{prime p}(1 - 1/p^k) = 1/zeta(k) which corresponds
to the known density of k-free numbers.
Let now S consists of 0 and squarefree numbers. Then (1)
should give the density of so-called exponentially squarefree
numbers (A209061).
On the other hand, Laszlo Toth in his Theorem 3 (see link in
A209061) obtained another formula for the same density
(without an explanation in detail (see proof)):
Prod_{prime p}(1+sum_{j>=4}((mu(j)^2-mu(j-1)^2)/p^j), (2)
where mu(n) is the Möbius function.
I asked Peter Moses to calculate this density by (1) and (2),
but he obtained different results: by (1), 0.955695..., while, by (2),
0.955923... . If anyone can verify these calculations?
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 04 September 2015 14:27
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Numbers with squarefree exponents in their prime factorizations
A better upper estimate for density of
A209061 is 1- sum_{k>=4}(1-|mu(k)|)*
(1/zeta(k+1) - 1/zeta(k))<0.95637
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 03 September 2015 18:20
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Numbers with squarefree exponents in their prime factorizations
An upper estimate for density of A209061
is 1-1/zeta(5)+1/zeta(4)=0.95955...
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Charles Greathouse [charles.greathouse at case.edu]
Sent: 02 September 2015 22:19
To: Sequence Fanatics Discussion list
Subject: [seqfan] Numbers with squarefree exponents in their prime factorizations
I was looking at sequence A209061 today and wondered what its density was.
With A046100 as a subsequence it's at least 1/zeta(4) = 0.92..., of course.
But I can't quite figure out what Euler product to use here -- can anyone
help out to improve the sequence?
Charles Greathouse
Analyst/Programmer
Case Western Reserve University
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