# Why is (pi(100)/1!+pi(pi(100))/2!+pi(pi(pi(100)))/3!+…)/100 so close to {log(10)}?

Consider the series defining the exponential function $\displaystyle{\sum_{k=0}^{\infty}\dfrac{x^{k}}{k!}}$ and replace $x^k$ by $f^{(k)}$ where $f^{(0)}=Id$ and $f^{(k+1)}=f\circ f^{(k)}$. Applying this to $\pi$, the counting function of primes, one gets the function $\displaystyle{E_{\pi}:x\mapsto\sum_{k=0}^{\infty}\dfrac{\pi^{(k)}(x)}{k!}}$.

When one tries to evaluate $\dfrac{E_{\pi}(100)-100}{100}$, one gets $0.30258333333...$, whereas $\{\log 10\}=0.30258509...$.

Is this just a coincidence or does something deeper lie under this numerical agreement?

I doubt anything deep lies under a formula containing $\pi(\pi(x))$ and higher compositions (even if one believes that the fractional part of the natural logarithm of something might be interesting). – Greg Martin Jun 9 '13 at 0:42
I would say it is no surprise that one can construct entirely unrelated expressions of that length whose values coincide up to something like 5 decimal digits. -- Just consider the number of possible such expressions and compare it with $10^5$ etc. – Stefan Kohl Jun 9 '13 at 9:36