The Hankel-transform relation for Rueppel complements and first differences

Counterexample and the correct signed determinant identity

Let

$$r_m=\begin{cases} 1,&m+1\text{ is a power of }2,\\ 0,&\text{otherwise}, \end{cases}$$

be the Rueppel sequence. Following Barry, Conjectures and results on some generalized Rueppel sequences, Conjecture 22, define

$$\begin{aligned} H_n&=\det\bigl(1-r_{i+j}\bigr)_{0\leq i,j\leq n},\\ h_n&=\det\bigl(r_{i+j+1}-r_{i+j}\bigr)_{0\leq i,j\leq n}. \end{aligned} \tag{1}$$

The proposed identity

$$|h_n|=\sqrt{|H_{n+1}|-|H_n|} \tag{2}$$

is false. In fact, the correct relationship is the signed identity

$$\boxed{ h_n^2 =(-1)^{\binom{n+1}{2}}H_n +(-1)^{\binom{n+2}{2}}H_{n+1} \qquad(n\geq0). } \tag{3}$$

Explicit counterexample

The relevant initial Rueppel values are

$$(r_0,r_1,r_2,r_3,r_4,r_5,r_6) =(1,1,0,1,0,0,0).$$

Consequently,

$$H_2= \det\begin{pmatrix} 0&0&1\\ 0&1&0\\ 1&0&1 \end{pmatrix} =-1, \qquad H_3= \det\begin{pmatrix} 0&0&1&0\\ 0&1&0&1\\ 1&0&1&1\\ 0&1&1&1 \end{pmatrix} =0,$$

whereas

$$h_2= \det\begin{pmatrix} 0&-1&1\\ -1&1&-1\\ 1&-1&0 \end{pmatrix} =1.$$

Thus

$$|H_3|-|H_2|=-1, \qquad |h_2|=1.$$

The right-hand side of (2) is therefore not even real. Equivalently, squaring the claimed identity would give the contradiction $1=-1$.

A universal determinant identity

The correction (3) follows from a more general identity. Let $(a_j)_{j\geq0}$ be any real sequence for which the Hankel determinants

$$\Delta_n=\det(a_{i+j})_{0\leq i,j\leq n}$$

are nonzero. Define

$$B_n=\det(1-a_{i+j})_{0\leq i,j\leq n}, \qquad D_n=\det(a_{i+j+1}-a_{i+j})_{0\leq i,j\leq n}.$$

Then, for every $n\geq0$,

$$\boxed{ D_n^2=(-1)^{n+1} \bigl(\Delta_n B_{n+1}+\Delta_{n+1}B_n\bigr). } \tag{4}$$

To prove this, write

$$A=(a_{i+j})_{0\leq i,j\leq n}, \qquad u=(1,\ldots,1)^{\mathsf T}, \qquad v=(a_{n+1},\ldots,a_{2n+1})^{\mathsf T},$$

and put

$$t=a_{2n+2}-v^{\mathsf T}A^{-1}v =\frac{\Delta_{n+1}}{\Delta_n}. \tag{5}$$

The matrix-determinant lemma gives

$$B_n=(-1)^{n+1}\Delta_n(1-K_n), \qquad K_n=u^{\mathsf T}A^{-1}u. \tag{6}$$

The block inverse of

$$A_{n+1}= \begin{pmatrix} A&v\\ v^{\mathsf T}&a_{2n+2} \end{pmatrix}$$

yields

$$K_{n+1}-K_n =\frac{(1-v^{\mathsf T}A^{-1}u)^2}{t}. \tag{7}$$

Finally, subtract consecutive columns, working from right to left, in the bordered determinant

$$\det\begin{pmatrix} A&v\\ u^{\mathsf T}&1 \end{pmatrix}.$$

Expansion along the last row gives

$$\Delta_n(1-v^{\mathsf T}A^{-1}u) =(-1)^{n+1}D_n. \tag{8}$$

Substituting (5) and (8) into (7),

$$K_{n+1}-K_n =\frac{D_n^2}{\Delta_n\Delta_{n+1}}. \tag{9}$$

Applying (6) at $n$ and $n+1$ and simplifying proves (4).

For the Rueppel sequence, the known Hankel evaluation recalled in Barry's introduction is

$$\Delta_n=(-1)^{\binom{n+1}{2}}. \tag{10}$$

Since

$$\Delta_{n+1}=(-1)^{n+1}\Delta_n,$$

specializing (4) to $a_j=r_j$ gives exactly (3). Hence the original absolute-value conjecture is disproved, and its replacement is an exact signed identity valid for every index.