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I have weird and vague question. We know the reciprocal of numbers

$$\sum_{k\leq n}\frac{1}{k}\sim \log n$$

and reciprocal of primes

$$\sum_{p\leq n}\frac{1}{p}\sim \log\log n$$

Now consider reciprocal of some sort of primes

$$\sum_{p*\leq n}\frac{1}{p*}\sim \log\log\log n$$

where $p*$ is the element of a subset of the prime number set. What would $p*$ be?

Thomas Andrews
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esege
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    Perhaps you can number the primes $1,2,3,\cdots$ and have $p$ be an element of you special subset iff its index is prime? This might not at all work, just an idea –  Feb 26 '16 at 13:14
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    First of all, there are many such subsets. You can algorithmically find a subset even so that $\sum_{p^\leq n} \frac{1}{p^{}} - \log\log\log n\to 0$, which is stronger than your statement. – Thomas Andrews Feb 26 '16 at 13:14
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    For example let $p_n^*$ be the closest prime to $n\log(n)\log(\log(n))$. Related question: http://math.stackexchange.com/questions/831283/nontrivial-subsequences-of-the-harmonic-series-that-diverge-on-the-order-of-lo – Winther Feb 26 '16 at 13:29
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    I think it's going to have to be denser than the set of primes with prime index: http://www.wolframalpha.com/input/?i=sum+1%2F(n+log(n)%5E2) – Dan Brumleve Feb 26 '16 at 15:47
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    btw if you want an explicit example then $p_{\lfloor n \log(\log(n)) \rfloor}$ for $n\geq 4$ should work. – Winther Feb 26 '16 at 16:18
  • Actually what i'm looking for is prime of to form of something like Mersenne primes, Wilson primes, primes of the form $1+n^2$ or something... – esege Feb 27 '16 at 00:13
  • @esege Fogetaboudit! It's not even known that there are infinitely many Mersenne primes, Wilson primes or primes on the form $n^2+1$ which is a requirement for the bounds you want. It's highly unlikely that you are going to find a natural subset of primes like the ones you mention satisfying the bound you want. – Winther Feb 27 '16 at 21:56

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