Talk:Harmonic series of the composites

Composite reciprocal constant

$C=\gamma -1-M=-0.684281547946...,\,$

A Google search of the following was NOT succesful

• Harmonic series of the composites (sum of the reciprocals of the composites)
• Composite reciprocal constant

A search on the OEIS (0.684281547946) or on Plouffe's Inverter of the following was NOT succesful

• 0.684281547946

Did I err in my reasoning about the composite reciprocal constant? — Daniel Forgues 00:43, 24 September 2012 (UTC)

Yes. You write

$\lim _{n\to \infty }\left(\sum _{\stackrel {c\leq n}{c{\rm {~composite}}}}{\frac {1}{c}}\right)\sim {\frac {\log n}{\log \log n}}.\,$

which I assume is intended to read

$\lim _{n\to \infty }{\frac {\log \log n}{\log n}}\sum _{\stackrel {c\leq n}{c{\rm {~composite}}}}{\frac {1}{c}}=1$

(since the n on the left is a dummy variable but the n on the right is free in your version) but in fact

$\lim _{n\to \infty }{\frac {\log \log n}{\log n}}\sum _{\stackrel {c\leq n}{c{\rm {~composite}}}}{\frac {1}{c}}=+\infty$

The correct version is

$\lim _{n\to \infty }{\frac {1}{\log n}}\sum _{\stackrel {c\leq n}{c{\rm {~composite}}}}{\frac {1}{c}}=1$

or, if you prefer,

$\sum _{\stackrel {c\leq n}{c{\rm {~composite}}}}{\frac {1}{c}}\sim \log n$

which is similar to the one for natural numbers:

$\sum _{k\leq n}{\frac {1}{k}}\sim \log n$

Charles R Greathouse IV 02:42, 24 September 2012 (UTC)

Oops, I really fumbled, I wrongly made the left side equal to the right side

$\log n-\log \log n\neq \left({\frac {\log n}{\log \log n}}\right),\,$

while it is obviously

$\log n-\log \log n=\log \left({\frac {n}{\log n}}\right),\,$

and

$\log n-\log \log n\sim \log n.\,$

I think this is right (is it?): the composite reciprocal constant is

$C=\gamma -1-M=-0.684281547946...,\,$

— Daniel Forgues 06:03, 24 September 2012 (UTC)

I want to be sure before creating the page composite reciprocal constant... (the search for −0.684281547946... was NOT succesful?!) — Daniel Forgues 06:34, 24 September 2012 (UTC)

Certainly you can partition the positive integers into 1, the primes, and the composites, so you get (sum over naturals) = (sum over primes) + (sum over composites) + 1. This does indeed give an asymptotic of $\log x-\log \log x+(\gamma -1-B_{1})+o(1),$ though I don't see any particular significance in this. (It's probably worth submitting a sequence for. I wouldn't write a wiki page but I'm sure you will.) I don't know off the top of my head what the right order for the error term should be; you could find it by summing the error terms from the other two (it can't be worse than that). I just wrote the conservative o(1) above since it's easy. Charles R Greathouse IV 17:49, 24 September 2012 (UTC)

I see that you used $\log x$ rather than $\log x-\log \log x$; this doesn't work. I'm correcting the page now. Charles R Greathouse IV 18:08, 24 September 2012 (UTC)

You're right, I was confusing the asymptotic behavior with the formula for the constant, where we have to subtract everything that is not $o(1)$. Thanks for your help! — Daniel Forgues 04:37, 25 September 2012 (UTC)

Since $C=\gamma -1-M=$ − 0.684281547946... (− A143524 − 1), I'm not sure if is worth submitting as a new sequence, or simply add a comment in A143524. — Daniel Forgues 04:37, 25 September 2012 (UTC)

Asymptotic behaviors

• The harmonic series of the integers has asymptotic growth $\scriptstyle \log x\,$
• The harmonic series of the primes has asymptotic growth $\scriptstyle \log \log x\,$
• The ??? has asymptic growth $\scriptstyle \log \log \log x\,$
• The ??? has asymptic growth $\scriptstyle \log \log \log \log x\,$
• The ??? has asymptic growth $\scriptstyle \log \log \log \log \log x\,$
• ...

— Daniel Forgues

You can pick any sequences which have growth ~ x, ~ x log x, ~ x log x log log x, ~ x log x log log x log log log x, .... Up to a constant factor you could take the next sequence to be good primes (A187094). Charles R Greathouse IV 23:06, 25 September 2012 (UTC)

For A187094 [Good primes (version 3)], the asymptotic behavior is $\scriptstyle a(n)\,\sim \,n\log n\log \log n/\log 2\,$. I'd like to find a sequence with asymptotic behavior $\scriptstyle a(n)\,\sim \,\log \log \log n\,$. — Daniel Forgues 01:58, 26 September 2012 (UTC)

I guess I could take the quotient of a sequence asymptotic to $\scriptstyle x\log x\log \log x\log \log \log x\,$ and a sequence asymptotic to $\scriptstyle x\log x\log \log x\,$ but that seems contrived. And Id' have to find two such sequences... — Daniel Forgues 01:58, 26 September 2012 (UTC)

Along your idea, I could take the quotient of the sequence of primes, asymptotic to $\scriptstyle n\log n\,$, and the sequence of positive integers, asymptotic to $\scriptstyle n\,$, to get a sequence of rationals asymptotic to $\scriptstyle \log n\,$.

$\{{\tfrac {2}{1}},{\tfrac {3}{2}},{\tfrac {5}{3}},{\tfrac {7}{4}},{\tfrac {11}{5}},{\tfrac {13}{6}},{\tfrac {17}{7}},{\tfrac {19}{8}},{\tfrac {23}{9}},{\tfrac {29}{10}},...\}\,$

— Daniel Forgues 02:11, 26 September 2012 (UTC)

I was giving sequences with the desired reciprocal sums. The sum of the reciprocals of the first n good(3) primes is approximately k log log log n where k = 1/log(2). If you wanted the sequence itself to have that behavior, just take a(n) = round(log log log n). Charles R Greathouse IV 03:49, 26 September 2012 (UTC)

Does this sequence converge or diverge? (I don't think its asymptotic growth is $\scriptstyle \log \log \log x\,$, I suspect that it converges.) The harmonic series of the superprime numbers (prime indexed primes) (sum of the reciprocals of the primeth primes $\scriptstyle p(p(n))\,$)

$S=\sum _{i=1}^{\infty }{\frac {1}{p(p(i))}}={\frac {1}{p(2)}}+{\frac {1}{p(3)}}+{\frac {1}{p(5)}}+{\frac {1}{p(7)}}+{\frac {1}{p(11)}}+{\frac {1}{p(13)}}+\cdots ={\tfrac {1}{3}}+{\tfrac {1}{5}}+{\tfrac {1}{11}}+{\tfrac {1}{17}}+{\tfrac {1}{31}}+{\tfrac {1}{41}}+{\tfrac {1}{59}}+{\tfrac {1}{67}}\cdots \,$

where $\scriptstyle p(n)\,$ is the $\scriptstyle n\,$-th prime. — Daniel Forgues 06:49, 27 September 2012 (UTC)

It converges. Charles R Greathouse IV 02:14, 28 September 2012 (UTC)

So, we have a whole sequence of constants as the sum of the harmonic series of superprimes with order of primeness at least $\scriptstyle k,\,k\,\geq \,2$.

$S_{k}=\sum _{i=1}^{\infty }{\frac {1}{\underbrace {p(\cdots (p(i))\cdots )} _{k{\rm {~times}}}}},\quad k\geq 2.\,$

— Daniel Forgues 03:41, 28 September 2012 (UTC)