The inverse power index lower-bound conjecture for the distribution $(0.75,0.25,0,\dots)$

For $n\ge 2$ voters, let $d$ be the desired power distribution defined by

$$d_1=0.75,\qquad d_2=0.25,\qquad d_i=0\quad\text{for }3\le i\le n.$$

For a simple game (including a complete simple game or a weighted voting game) $\chi$, let $\mathcal{SS}(\chi)$ and $\mathcal{BZ}(\chi)$ denote its Shapley–Shubik and Banzhaf power vectors, respectively, and let $k_m$ and $l_m$ be defined by $k_1=1$, $l_1=2$, $l_2=5$, and

$$k_m=\begin{cases}2k_{m-1}&m\equiv0\pmod 2,\\8k_{m-1}-1&m\equiv1\pmod 2,\end{cases}\qquad m\ge2,$$ $$l_m=\begin{cases}2l_{m-1}+3&m\equiv0\pmod 2,\\8l_{m-1}-2&m\equiv1\pmod 2,\end{cases}\qquad m\ge3.$$

Inverse power index lower-bound conjecture. Every such simple game satisfies

$$\left\|\mathcal{SS}(\chi)-d\right\|_1\ge\frac13$$

and

$$\left\|\mathcal{BZ}(\chi)-d\right\|_1\ge\frac{k_{\left\lceil n/2\right\rceil}}{l_{\left\lceil n/2\right\rceil}}.$$

The distribution $(0.75,0.25,0,\dots)$ is motivated by its maximal $\ell^1$-distance from the achievable two-voter power vectors for both indices. Computations for games with at most $16$ voters support the recursive Banzhaf bound, but the paper gives no proof of the conjectured inequalities.