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

Proof of the period-four Rueppel Hankel conjecture

Let

$$r(x)=\sum_{j\geq0}r_jx^j =\sum_{k\geq0}x^{2^k-1}$$

be the Rueppel generating function, and write

$$F(x)=1-\frac{x}{r(x^2)}=\sum_{j\geq0}f_jx^j, \qquad T_n=\det(f_{i+j})_{0\leq i,j\leq n}.$$

We prove that

$$\boxed{ T_{4j}=1, \qquad T_{4j+1}=-1, \qquad T_{4j+2}=-1, \qquad T_{4j+3}=0 \quad(j\geq0). } \tag{1}$$

This is Conjecture 10 in Barry, Conjectures and results on some generalized Rueppel sequences.

Reciprocal Hankel identities

Let

$$A(x)=\sum_{j\geq0}a_jx^j, \qquad a_0=1, \qquad Q(x)=\frac1{A(x)}=\sum_{j\geq0}q_jx^j.$$

For $m\geq1$, formal-series inversion gives

$$\begin{aligned} \det(q_{i+j+1})_{0\leq i,j<m} &=(-1)^m\det(a_{i+j+1})_{0\leq i,j<m},\\ \det(q_{i+j})_{0\leq i,j<m} &=(-1)^{m-1}\det(a_{i+j+2})_{0\leq i,j<m-1}. \end{aligned} \tag{2}$$

Here the determinant of a zero-by-zero matrix is $1$.

For completeness, let $L=(a_{i-j})_{0\leq j\leq i<m}$ be the unit lower triangular Toeplitz matrix, and let $U=(q_{j-i})_{0\leq i\leq j<m}$ be its unit upper triangular counterpart. The convolution identity

$$\sum_{k=0}^{d}a_{d-k}q_k=0 \qquad(d\geq1) \tag{3}$$

implies

$$L(q_{i+j+1})_{i,j<m} =-(a_{i+j+1})_{i,j<m}U,$$

which proves the first equality in (2). Similarly, $L(q_{i+j})_{i,j<m}$ has first column $(1,0,\ldots,0)^{\mathsf T}$. Deleting that column and the first row leaves

$$-(a_{i+j+2})_{0\leq i,j<m-1}U', \qquad U'=(q_{j-i})_{0\leq i\leq j<m-1},$$

proving the second equality in (2).

Shifted Rueppel determinants

Set

$$\begin{aligned} R_m&=\det(r_{i+j})_{0\leq i,j<m},\\ J_m&=\det(r_{i+j+1})_{0\leq i,j<m},\\ L_m&=\det(r_{i+j+2})_{0\leq i,j<m}, \end{aligned}$$

with $R_0=J_0=L_0=1$. The standard Rueppel determinant evaluation, recalled in Barry's introduction, is

$$R_m=(-1)^{\binom m2}. \tag{4}$$

The Rueppel coefficients satisfy

$$r_{2j}=\begin{cases}1,&j=0,\\0,&j\geq1,\end{cases} \qquad r_{2j+1}=r_j. \tag{5}$$

Reordering both rows and columns by parity in the matrix for $J_m$, its off-diagonal parity blocks vanish. Its diagonal blocks are ordinary and once-shifted Rueppel Hankel matrices. Hence

$$J_{2t}=R_tJ_t, \qquad J_{2t+1}=R_{t+1}J_t. \tag{6}$$

It follows inductively from (4) that

$$J_m\in\{-1,1\} \qquad(m\geq0). \tag{7}$$

For $L_m$, the diagonal parity blocks vanish. Thus an odd-size matrix has determinant zero, whereas an even-size matrix has two identical once-shifted blocks:

$$L_{2t+1}=0, \qquad L_{2t}=(-1)^tJ_t^2=(-1)^t. \tag{8}$$

Now specialize (2) to $A(x)=r(x)$. Equations (7) and (8) give

$$\left|\det(q_{i+j+1})_{0\leq i,j<m}\right|=1, \tag{9}$$

and

$$\left|\det(q_{i+j})_{0\leq i,j<m}\right| = \begin{cases} 1,&m\text{ odd},\\ 0,&m\text{ even}. \end{cases} \tag{10}$$

The target Hankel determinants

Since

$$F(x)=1-xQ(x^2),$$

its coefficients satisfy

$$f_0=1, \qquad f_{2j}=0\quad(j\geq1), \qquad f_{2j+1}=-q_j. \tag{11}$$

Reorder the rows and columns of its Hankel matrix by parity. For an odd-size matrix, expansion along the unique nonzero diagonal-parity entry gives

$$T_{2m}=(-1)^m \det(q_{i+j+1})_{0\leq i,j<m}^{\,2} =(-1)^m. \tag{12}$$

For an even-size matrix, the two off-diagonal blocks are square, giving

$$T_{2m+1}=(-1)^{m+1} \det(q_{i+j})_{0\leq i,j\leq m}^{\,2} = \begin{cases} -1,&m\text{ even},\\ 0,&m\text{ odd}. \end{cases} \tag{13}$$

Combining (12) and (13) proves the full period-four formula (1), namely

$$(T_n)_{n\geq0} =(1,-1,-1,0,1,-1,-1,0,\ldots).$$