# Solving $f(\log x)$

A generalization of the conjecture

$$\pi(x+x^{\theta}) - \pi(x) \sim \frac{x^\theta}{\log x}$$ (Ingham, 1937 or earlier) might be

$$\Delta \pi_k = \pi_k((x+1)^2) - \pi_k(x^2)\sim \frac{x}{\log x}\frac{(\log\log x^2)^{(k-1)}}{(k-1)!}$$

in which $\pi_k(x)$ is the number of numbers with k primes including repetitions not exceeding x.

For the case $k = 1$ we have $\Delta \pi_1 \sim \frac{x}{\log x},$ suggesting that the number of primes on a square interval approaches that of the interval $[1,x].$ All very speculative. My question is simply whether we can find a similar (speculative) statement for $\Delta \pi_2,$ etc., relating the number of primes on the square interval to that of some interval $[1,x]?$ In other words, can we solve for example:

$$\frac{x}{\log x}\log\log (x^2)= \frac{y}{\log y} \sim \pi(y)$$ for y in terms of x?

It's easy enough to plot $f(x)=\frac{x}{\log x}\log\log x$ and find $\pi(y)\approx f(x).$ For example:

$$\frac{1000}{\log 1000}\log\log 1000^2 \approx \frac{3048}{\log 3048}.$$

Thanks for any insight.

I have failed the Turing test twice today, so editing appreciated.

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The solution $> e$ of the equation $y/\log(y) = z$ is $y = - z\; \rm{LambertW}(-1,-1/z)$ (in Maple's terminology) for $z > e$.

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Mathematica did something like this for z = y/log(y), but did not like stuff like f(z) = y/log(y). So maybe that is asking too much? If so, thanks and I will accept this. – daniel Jan 6 '13 at 20:45
If you can solve $s = y/\log(y)$ for $y$ when $s > e$, and $f(z) > e$ for sufficiently large $z$, just plug in $s = f(z)$. – Robert Israel Jan 6 '13 at 20:51
OK will try that. – daniel Jan 6 '13 at 20:54