[seqfan] A262373
Moses, Peter J. C.
mows at mopar.freeserve.co.uk
Thu Sep 24 14:13:32 CEST 2015
Hi Vladimir,
Happened to take a quick look at A262373.
You have a(46) = 97. This cannot be as 97 has appeared at a(30)
a(46) = 107.
I get:
{2, 5, 3, 13, 11, 31, 23, 43, 53, 73, 7, 17, 41, 61, 71, 37, 83, 103, 101,
131, 151, 181, 191, 211, 241, 251, 271, 47, 67, 97, 19, 281, 311, 113, 331,
163, 401, 421, 431, 173, 461, 491, 29, 59, 79, 107, 521, 541, 571, 127, 601,
631, 193, 641, 661, 691, 89, 109, 701, 137, 751, 157, 761, 167, 811, 821,
881, 911, 139, 941, 149, 971, 179, 991, 199, 1021, 1031, 1051, 1061, 1091,
1151, 1171, 1181, 1201, 1231, 1291, 1301, 1321, 1361, 1381, 1451, 1471,
1481, 1511, 1531, 1571, 1601, 1621, 1721, 1741, 1801, 1811, 1831, 1861,
1871, 1901, 1931, 1951, 2011, 2081, 2111, 2131, 2141, 2161, 2221, 2251,
2281, 2311, 223, 233, 263, 283, ...}
Or for OEIS DATA:
2, 5, 3, 13, 11, 31, 23, 43, 53, 73, 7, 17, 41, 61, 71, 37, 83, 103, 101,
131, 151, 181, 191, 211, 241, 251, 271, 47, 67, 97, 19, 281, 311, 113, 331,
163, 401, 421, 431, 173, 461, 491, 29, 59, 79, 107, 521, 541, 571, 127, 601,
631, 193, 641, 661, 691, 89, 109
Best regards,
Peter.
--------------------------------------------------
From: "Vladimir Shevelev" <shevelev at bgu.ac.il>
Sent: Friday, September 18, 2015 8:25 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Numberswith squarefree exponents in their primefactorizations> 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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>
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> _______________________________________________
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> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
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