The modulo-2 Hankel transform of $1-x+x^2/r(x^2)$

Proof of the period-eight Rueppel Hankel conjecture

Let

$$r(x)=\sum_{j\geq 0}x^{2^j-1}, \qquad F(x)=1-x+\frac{x^2}{r(x^2)} =\sum_{j\geq0}f_jx^j,$$

and write

$$h_n=\det(f_{i+j})_{0\leq i,j\leq n}.$$

We prove, for every $n\geq0$, that

$$\boxed{ h_n\equiv \begin{cases} 0\pmod 2,&n\equiv1,6\pmod8,\\ 1\pmod 2,&n\equiv0,2,3,4,5,7\pmod8. \end{cases}} \tag{1}$$

This establishes Conjecture 11 of Paul Barry, Conjectures and results on some generalized Rueppel sequences.

Reduction to Catalan moments

Let

$$C(x)=\sum_{j\geq0}C_jx^j, \qquad C_j=\frac1{j+1}\binom{2j}{j},$$

be the Catalan generating function. Its defining identity is

$$C(x)=1+xC(x)^2. \tag{2}$$

On the other hand,

$$r(x)=1+xr(x^2). \tag{3}$$

Over $\mathbb F_2[[x]]$, Frobenius gives $r(x^2)=r(x)^2$. Thus (2) and (3), together with their common constant coefficient, imply

$$r(x)\equiv C(x)\pmod2. \tag{4}$$

Because both series have constant coefficient $1$, inversion preserves this congruence. Consequently $F$ is coefficientwise congruent to the integral series

$$\widetilde F(x) =1-x+\frac{x^2}{C(x^2)}. \tag{5}$$

By (2),

$$Q(x)=\frac1{C(x)}=1-xC(x)=\sum_{j\geq0}q_jx^j, \tag{6}$$

so

$$q_0=1, \qquad q_j=-C_{j-1}\quad(j\geq1). \tag{7}$$

It therefore suffices to evaluate the Hankel determinants of

$$\widetilde F(x)=1-x+x^2Q(x^2) \tag{8}$$

over the integers and then reduce the result modulo $2$.

An auxiliary Catalan determinant

For $k\geq1$, let $A_k$ be the size-$k$ Hankel determinant of

$$1+xQ(x)=1+x-x^2C(x).$$

We claim that

$$A_k=(-1)^{k-1} \left(1+\sum_{j=1}^{k-1}j^2\right). \tag{9}$$

To see this, put $m=k-1$ and introduce the Catalan moment matrix

$$M_m=(C_{i+j})_{0\leq i,j<m}.$$

Its standard ballot-triangle factorization is

$$M_m=L_mL_m^{\mathsf T}, \qquad L_m=(\ell_{i,j})_{0\leq i,j<m}, \qquad \ell_{i,j} =\binom{2i}{i-j}-\binom{2i}{i-j-1} \quad(0\leq j\leq i). \tag{10}$$

Here out-of-range binomial coefficients are zero, and $L_m$ is unit lower triangular. Define

$$v=(1,-C_0,-C_1,\ldots,-C_{m-2})^{\mathsf T}, \qquad z_j=(-1)^j(j+1). \tag{11}$$

The ballot generating function

$$\sum_{i\geq j}\ell_{i,j}x^i =x^jC(x)^{2j+1} \tag{12}$$

shows that

$$\begin{aligned} \sum_{i\geq0} \left(\sum_{j=0}^{i} (-1)^j(j+1) \left[ \binom{2i}{i-j}-\binom{2i}{i-j-1} \right] \right)x^i &=\frac{C(x)}{(1+xC(x)^2)^2}\\ &=\frac1{C(x)}\\ &=1-xC(x). \end{aligned} \tag{13}$$

Hence $L_mz=v$. The Hankel matrix defining $A_k$ has the block form

$$\begin{pmatrix} 1&v^{\mathsf T}\\ v&-M_m \end{pmatrix}.$$

Since $\det M_m=1$, its Schur complement gives

$$\begin{aligned} A_k &=(-1)^m\left(1+v^{\mathsf T}M_m^{-1}v\right)\\ &=(-1)^m\left(1+z^{\mathsf T}z\right)\\ &=(-1)^m\left(1+\sum_{j=1}^{m}j^2\right), \end{aligned}$$

which proves (9).

The usual shifted Catalan Hankel evaluations are

$$\det(C_{i+j})_{0\leq i,j<t} =\det(C_{i+j+1})_{0\leq i,j<t}=1, \qquad \det(C_{i+j+2})_{0\leq i,j<t}=t+1. \tag{14}$$

The first two identities follow from their respective unit ballot-triangle factorizations; the third follows from the first two by the Desnanot--Jacobi determinant identity.

For any series $P(x)=\sum_{j\geq0}p_jx^j$ with $p_0=1$ and reciprocal $P(x)^{-1}=\sum_{j\geq0}b_jx^j$, formal-series inversion gives

$$\det(b_{i+j})_{0\leq i,j<t} =(-1)^{t-1} \det(p_{i+j+2})_{0\leq i,j<t-1}. \tag{15}$$

Applying this to $P=C$, equations (7), (14), and (15) yield

$$\begin{aligned} B_t &:=\det(q_{i+j})_{0\leq i,j<t} =(-1)^{t-1}t,\\ D_t &:=\det(q_{i+j+1})_{0\leq i,j<t} =(-1)^t,\\ E_t &:=\det(q_{i+j+2})_{0\leq i,j<t} =(-1)^t. \end{aligned} \tag{16}$$

Every determinant of size zero is understood to be $1$.

The complete determinant evaluation

Let

$$\widetilde H_N =\det(\widetilde f_{i+j})_{0\leq i,j<N}, \qquad \widetilde F(x)=\sum_{j\geq0}\widetilde f_jx^j.$$

By (8),

$$\widetilde f_0=1, \qquad \widetilde f_1=-1, \qquad \widetilde f_{2j+2}=q_j, \qquad \widetilde f_{2j+3}=0. \tag{17}$$

Reorder both rows and columns by even and odd indices. For

$$e=\left\lceil\frac N2\right\rceil, \qquad o=\left\lfloor\frac N2\right\rfloor,$$

the even-even block is the size-$e$ matrix defining $A_e$, the odd-odd block is the size-$o$ matrix defining $B_o$, and both off-diagonal blocks have their only nonzero entry, $-1$, in the upper-left corner. Expansion in these two entries gives

$$\widetilde H_N=A_eB_o-D_{e-1}E_{o-1} \qquad(N\geq2). \tag{18}$$

Substitution from (9) and (16) yields the stronger exact formulas

$$\boxed{ \begin{aligned} \widetilde H_{2m} &=m\left(1+\sum_{j=1}^{m-1}j^2\right)-1,\\ \widetilde H_{2m+1} &=1-m\left(1+\sum_{j=1}^{m}j^2\right) \end{aligned} \qquad(m\geq1),} \tag{19}$$

together with $\widetilde H_1=1$.

Since $j^2\equiv j\pmod2$, we have

$$\sum_{j=1}^{t}j^2\equiv \left\lceil\frac t2\right\rceil \pmod2. \tag{20}$$

Equations (19)--(20) now show that, by matrix size $N$, the determinant parities are

$$(\widetilde H_N\bmod2)_{N\geq1} =\overline{1,0,1,1,1,1,0,1}. \tag{21}$$

Finally, (4)--(5) give

$$h_n\equiv\widetilde H_{n+1}\pmod2,$$

so (21) proves the asserted period-eight pattern (1) for every index.