# Identification of a curious function

During computation of some Shapley values (details below), I encountered the following function: $$f\left(\sum_{k \geq 0} 2^{-p_k}\right) = \sum_{k \geq 0} \frac{1}{(p_k+1)\binom{p_k}{k}},$$ where $$p_0 > 0$$ and $$p_{k+1} > p_k$$ for all $$k$$. In other words, the input to $$f$$ is the binary expansion of a real number in the range $$[0,1]$$, and the $$p_k$$ correspond to the positions of $$1$$s in the binary expansion.

For example, $$f(2^{-t}) = 1/(t+1)$$, so $$f(1/2) = 1/2$$, $$f(1/4) = 1/3$$ and so on. More complicated examples are $$f(5/8) = f(2^{-1} + 2^{-3}) = 1/2 + 1/(4\cdot 3) = 7/12$$ and $$f(2/3) = f\left(\sum_{k \geq 0}2^{-(2k+1)}\right) = \sum_{k \geq 0} \frac{1}{(2k+2)\binom{2k+1}{k}} = \frac{\pi}{\sqrt{27}}.$$

The function $$f$$ is a continuous increasing function satisfying $$f(0) = 0$$, $$f(1) = 1$$, and $$f(1-t) = 1-f(t)$$ for $$t \in [0,1]$$. It has vertical asymptotes at dyadic points.

Here is a plot of $$f$$:

Is the function $$f$$ known?

Here is where $$f$$ came from. Let $$n \geq 1$$ be an integer and let $$t \in [0,1]$$. For a permutation $$\pi$$ of the numbers $$\{ 2^{-m} : 0 \leq m \leq n-1 \}$$ satisfying $$\pi^{-1}(1) = i$$, we say that $$\pi$$ is pivotal if $$\sum_{j. Let $$f_n(t)$$ be the probability that a random $$\pi$$ is pivotal. Then $$f(t) = \lim_{n \rightarrow \infty} f_n(t)$$.

For example, take $$n = 4$$. The permutation $$1/8,1/2,1,1/4$$ is pivotal for $$t \in (5/8,1]$$. For all $$n \geq 2$$ we have $$f_n(1/2) = 1/2$$, since $$\pi$$ is pivotal iff $$1$$ appears before $$1/2$$ in $$\pi$$. The general formula for $$f$$ is derived in a similar way.

We leave it to the reader to figure out how $$f_n$$ measures some Shapley value. The functions $$f_n$$ are step functions with steps of length $$1/2^{n-1}$$. They are left-continuous, and are equal to $$f$$ at the breakpoints.

This question was also asked on MO.

• It seems spiritually related to the inverse of Minkowski's question mark function.
– user856
Aug 18, 2013 at 8:06
• I get $[f(x)-1/2] = [f(x/2)-1/2] + [f((x+1)/2)-1/2]$. Aug 18, 2013 at 21:16
• @BrunoJoyal Since $f(t) + f(1-t) = 1$, the integral is $\int_0^1 f(t) dt = 1/2$. Apr 28, 2014 at 0:54
• @Yuval: Based on the evidence, I conjecture that the higher moments are also rational. (I'm joking, but due to the apparent high degree of symmetry of your function, part of me wouldn't even be surprised - and the rest of me would be delighted!) Apr 28, 2014 at 1:12
• @Yuval, I saw something similar when reviewing election strategies (as in Politics) in an election where people vote for a list of candidates, there is a number of places to be elected $N$ (say a senate), and lists get as many places as percentage votes won: $places = floor(N P_l)$. The tricky part was that there is a remanent after allocating all lists, as one cannot allocate a fraction of a place, an allocation then was from greatest residual to lowest. So, what's the best strategy, split the party in many lists or nominate a single one? Plotting, we arrive at something similar. Jun 2, 2014 at 19:30