# Identification of a curious function

During computation of some Shapley values (details below), I encountered the following function:
$$f\left(\sum_{k \geq 0} 2^{-p_k}\right) = \sum_{k \geq 0} \frac{1}{(p_k+1)\binom{p_k}{k}}, f\left(\sum_{k \geq 0} 2^{-p_k}\right) = \sum_{k \geq 0} \frac{1}{(p_k+1)\binom{p_k}{k}},$$
where $$p_0 > 0p_0 > 0$$ and $$p_{k+1} > p_kp_{k+1} > p_k$$ for all $$kk$$. In other words, the input to $$ff$$ is the binary expansion of a real number in the range $$[0,1][0,1]$$, and the $$p_kp_k$$ correspond to the positions of $$11$$s in the binary expansion.

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

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

Here is a plot of $$ff$$:

Is the function $$ff$$ known?

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

For example, take $$n = 4n = 4$$. The permutation $$1/8,1/2,1,1/41/8,1/2,1,1/4$$ is pivotal for $$t \in (5/8,1]t \in (5/8,1]$$. For all $$n \geq 2n \geq 2$$ we have $$f_n(1/2) = 1/2f_n(1/2) = 1/2$$, since $$\pi\pi$$ is pivotal iff $$11$$ appears before $$1/21/2$$ in $$\pi\pi$$. The general formula for $$ff$$ is derived in a similar way.

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

This question was also asked on MO.