# [seqfan] Re: Numbers with squarefree exponents in their prime factorizations

Vladimir Shevelev shevelev at bgu.ac.il
Tue Sep 15 19:45:54 CEST 2015

```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,

________________________________________
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,

________________________________________
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,

________________________________________
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,

________________________________________
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,
________________________________________
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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```