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I am trying to figure out if for each $n$ we have

$$\sum_{i=1}^n\pi (2^i)+(1-1/2^n)\Bigg(\sum_{i=1}^{\lfloor n Log2/Log3\rfloor-1}\pi(3^i)+\sum_{i=1}^{\lfloor n log2/log5\rfloor-1}\pi(5^i)+\cdots+\sum_{i=1}^{\lfloor n Log2/LogP\rfloor-1}\pi(P^i)\Bigg) >2^{n-1}-2+\pi(2^n)(1+\pi(2^n)/2,$$ where $\pi(x)$ is

PrimePi[x]

and $P$ is the largest prime less than $2^n$, which is approximately

PrimePi[2^n] Log[PrimePi[2^n]].

I'm not sure how to code this since I want the sum of sums in the large parenthesis to be only over the primes from 3 to $P$, but here's some of the code:

Sum[PrimePi[2^i],{i,1,n}]+(1-1/2^n){ Sum[PrimePi[3^i],{i,1,Floor[n Log2/Log3]-1}]+ 

Sum[PrimePi[5^i],{i,1,Floor[n Log2/Log3]-1}] +...+Sum[PrimePi[P^i],{i,1,Floor[n Log2/LogP]-1}]}
> 2^(n-1)-2+PrimePi[2^i](1+PrimePi[2^i])/2

Can Mathematica determine whether this inequality is true? If not, will it list a plot for values of $n$?

It might be more convenient to approximate

PrimePi[x] 

with

 x/Log[x]

When making this substitution, Wolframalpha https://www.wolframalpha.com/input/?i=sum+i%3D1+to+(Floor%5Bnlog+2%2Fln3%5D)+3%5Ei%2F(i+ln3) gives a way to rewrite the sums, e.g., we can replace

Sum[PrimePi[3^i],{i,1,Floor[n Log2/Log3]-1}

by

((-I) (Pi - I 3^(1 +
Floor[(n Log[2])/Log[3]]) LerchPhi[3, 1, 1 + Floor[(n Log[2])/Log[3]]]
- I Log[2]))/Log[3].
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  • $\begingroup$ There are a few typos in your post -- eg, i on the r.h.s. If I've translated correctly, are you looking for something like this: Table[ P = NextPrime[2^n, -1]; Sum[PrimePi[2^i], {i, 1, n}] + (1 - 1/2^n) Sum[ Sum[PrimePi[j^i], {i, 1, Floor@N[n Log[2]/Log[j]] - 1}], {j, 3, P}], {n, 2, 10}] $\endgroup$ – aardvark2012 Jun 29 '17 at 11:54
  • $\begingroup$ I would suggest replacing all the "hard" functions with approximations that are easy to compute, check the resulting inequality, then try to reason about the error terms resulting from the approximation. Only if that fails would one want to get more refined. (Maybe you already did this, just wanted to mention the possibility in case not.) $\endgroup$ – Daniel Lichtblau Jun 29 '17 at 17:10

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