The A268411 product formula for Rueppel Hankel transforms

Proof

Let

$$h_n=\det(r_{i+j})_{0\leq i,j\leq n}, \qquad H_n=\det(r_{i+j+1})_{0\leq i,j\leq n},$$

where $r_j=1$ exactly when $j=2^a-1$ for some $a\geq0$, and $r_j=0$ otherwise.

The shifted Rueppel sequence is an aeration of the original sequence:

$$r_{2j+1}=r_j, \qquad r_{2j+2}=0.$$

Order the rows and columns of the matrix defining $H_n$ by parity. Because all mixed-parity entries vanish, the matrix becomes block diagonal. For $n=2m$, its two blocks have determinants $h_m$ and $H_{m-1}$; for $n=2m+1$, they have determinants $h_m$ and $H_m$. Hence

$$H_0=h_0=1,$$ $$H_{2m}=H_{m-1}h_m \quad (m\geq1), \qquad H_{2m+1}=H_mh_m \quad (m\geq0). \qquad (1)$$

The ordinary Rueppel Hankel determinants satisfy

$$h_n=(-1)^{\binom{n+1}{2}}. \qquad (2)$$

This is also the evaluation recorded with the aeration recurrence in Proposition 4 of:

https://arxiv.org/abs/2005.04066

Put $\epsilon_m=(-1)^m$ and

$$s_n=(-1)^nh_nH_n.$$

Equation (2) gives

$$h_{2m}=\epsilon_m, \qquad h_{2m+1}=\epsilon_{m+1}, \qquad h_m=\epsilon_mh_{m-1}\quad(m\geq1).$$

Substitution into (1) now gives

$$s_0=1, \qquad s_{2m+1}=s_m, \qquad s_{2m+2}=\epsilon_ms_m. \qquad (3)$$

Let $R(u)$ denote the number of maximal runs of $1$'s in the ordinary binary expansion of $u$, with $R(0)=0$. Appending a zero preserves the number of runs, while appending a one creates a new run exactly when the preceding number is even. Thus

$$R(2u)=R(u),$$

and

$$R(2u+1)= \begin{cases} R(u)+1,&u\text{ even},\\ R(u),&u\text{ odd}. \end{cases}$$

Define

$$q_n=(-1)^{R(n+1)+1}.$$

Then $q_0=1$. Applying the preceding binary recurrences to $2m+2=2(m+1)$ and $2m+3=2(m+1)+1$ gives

$$q_{2m+1}=q_m, \qquad q_{2m+2}=\epsilon_mq_m. \qquad (4)$$

Equations (3) and (4) give the same initial value and reduce every positive index to a smaller one. Strong induction therefore yields

$$s_n=q_n$$

for every $n\geq0$. Consequently,

$$\frac{1+(-1)^nh_nH_n}{2} = \frac{1+(-1)^{R(n+1)+1}}{2} = R(n+1)\pmod 2.$$

The right-hand side is exactly $\operatorname{A268411}(n+1)$. This proves Conjecture 23.

Conjecture source: https://arxiv.org/abs/2107.00442