The A005811 Hankel transform for $1-x+x^2/(1+x^2r(x^2))$

Counterexample and the corrected binary-run Hankel theorem

Let

$$r(x)=\sum_{j\geq0}x^{2^j-1}, \qquad F_{\pm}(x)=1-x+\frac{x^2}{1\pm x^2r(x^2)},$$

and define

$$T_n^{\pm} =\det\bigl([x^{i+j}]F_{\pm}(x)\bigr)_{0\leq i,j\leq n}.$$

The literal statement

$$T_n^+=\operatorname{A005811}(n-1) \tag{1}$$

is false. Indeed,

$$F_+(x)=1-x+x^2-x^4+O(x^6),$$

so at $n=2$ its Hankel determinant is

$$T_2^+ =\det\begin{pmatrix} 1&-1&1\\ -1&1&0\\ 1&0&-1 \end{pmatrix} =-1, \qquad \operatorname{A005811}(1)=1. \tag{2}$$

Thus $T_2^+\neq\operatorname{A005811}(1)$. The issue is a missing absolute value, not a failure of the intended binary-run pattern. The source's own displayed Hankel sequence already contains negative terms. Moreover, its asserted offset is undefined at $n=0$, so that initial case must be handled separately.

Write $\rho(m)$ for the number of runs in the ordinary binary expansion of $m\geq1$, with $\rho(0)=0$, and put

$$J_m=\det(r_{i+j+1})_{0\leq i,j<m}, \qquad J_0=1. \tag{3}$$

We prove the following exact signed formulas:

$$\boxed{ \begin{aligned} T_0^+&=T_0^-=1,\\ T_1^+&=T_1^-=0,\\ T_n^+&=(-1)^{n-1}J_{n-1}\rho(n-1),\\ T_n^-&=-J_{n-1}\rho(n-1) \qquad(n\geq2). \end{aligned}} \tag{4}$$

In particular, because $J_m\in\{-1,1\}$,

$$\boxed{ |T_n^+|=|T_n^-| =\rho(n-1) =\operatorname{A005811}(n-1) \qquad(n\geq2). } \tag{5}$$

Consequently, the literal Conjecture 15 in Paul Barry, Conjectures and results on some generalized Rueppel sequences requires an absolute value; (5) proves its corrected intended binary-run statement for every meaningful index. It simultaneously covers the denominator with a plus sign appearing in the conjecture and the denominator with a minus sign underlying its displayed initial example.

A universal two-sign Hankel identity

For a formal power series $U$, write

$$\mathcal H_N(U) =\det\bigl([x^{i+j}]U(x)\bigr)_{0\leq i,j<N}, \qquad \mathcal H_0(U)=1.$$

Let $A(x)$ be any formal power series with $A(0)=1$, and set

$$\begin{aligned} G(x)&=x+\frac1{A(x^2)},\\ F_{\pm}(x)&=1-x+\frac{x^2}{1\pm x^2A(x^2)}. \end{aligned} \tag{6}$$

We claim that, for every $N\geq3$,

$$\boxed{ \begin{aligned} \mathcal H_N(F_+)&=-\mathcal H_{N-2}(G),\\ \mathcal H_N(F_-)&=(-1)^{N-2}\mathcal H_{N-2}(G). \end{aligned}} \tag{7}$$

We use the standard Hankel continued-fraction transformation

$$U(x)=\frac{x^d}{1+u(x)x-x^{d+2}W(x)} \quad\Longrightarrow\quad \mathcal H_N(U) =(-1)^{\binom{d+1}{2}}\mathcal H_{N-d-1}(W), \tag{8}$$

where $d\geq0$, $\deg u\leq d$, and $N\geq d+1$. This is Lemma 36 of J.-P. Allouche, G.-N. Han, and J. Shallit, On some conjectures of P. Barry, Journal of Number Theory 228 (2021), 108--132; their preprint gives the same identity.

Define

$$V(x)=\frac{1-x-G(x)^{-1}}{x^2}, \qquad W_{\pm}(x)=\frac{1+x-F_{\pm}(x)^{-1}}{x^2}. \tag{9}$$

These are formal power series because $A(0)=1$. Straightforward substitution from (6) gives the exact identities

$$\begin{aligned} G(x)&=\frac1{1-x-x^2V(x)},\\ F_{\pm}(x)&=\frac1{1+x-x^2W_{\pm}(x)},\\ W_+(x)&=\frac{x}{1+2x^2-x^3V(-x)},\\ W_-(x)&=\frac{x}{1-2x+2x^2+x^3V(x)}. \end{aligned} \tag{10}$$

Applying (8), first with $d=0$ and then with $d=1$, yields

$$\begin{aligned} \mathcal H_N(F_+) &=\mathcal H_{N-1}(W_+)\\ &=-\mathcal H_{N-3}(V(-x)),\\ \mathcal H_N(F_-) &=\mathcal H_{N-1}(W_-)\\ &=-\mathcal H_{N-3}(-V(x)). \end{aligned} \tag{11}$$

For every formal series $V$, diagonal conjugation and scalar multiplication give

$$\mathcal H_j(V(-x))=\mathcal H_j(V(x)), \qquad \mathcal H_j(-V(x))=(-1)^j\mathcal H_j(V(x)). \tag{12}$$

Finally, the first equation in (10) and identity (8) imply

$$\mathcal H_{N-2}(G)=\mathcal H_{N-3}(V). \tag{13}$$

Equations (11)--(13) prove both identities in (7), without any assumption on $A$ beyond its constant coefficient.

Evaluation in the Rueppel case

Now take $A=r$, and let

$$Q(x)=\frac1{r(x)}=\sum_{j\geq0}q_jx^j.$$

Since

$$r(x)=1+xr(x^2),$$

we have

$$\frac1{G(x)}=\frac{1-Q(x)}x =1-x-x^2\sum_{j\geq0}q_{j+3}x^j. \tag{14}$$

The $d=0$ case of (8) therefore gives

$$\mathcal H_k(G)=\det(q_{i+j+3})_{0\leq i,j<k-1}. \tag{15}$$

The rectangular complementary-minor identity for reciprocal series is

$$\det(q_{i+j+3})_{0\leq i,j<t} =(-1)^{t+1} \det(r_{i+j-1})_{0\leq i,j<t+2}, \qquad r_{-1}=0. \tag{16}$$

It follows, for example, by applying the dual Jacobi--Trudi identity to the rectangular partition $(t^{\,t+2})$. Thus

$$\mathcal H_k(G)=(-1)^k\mathcal H_{k+1}(xr). \tag{17}$$

Let

$$D_m=\mathcal H_m(1-xr(x)).$$

Changing only the upper-left entry in the corresponding Hankel matrix gives

$$D_m=(-1)^m\mathcal H_m(xr)+(-1)^{m-1}J_{m-1}. \tag{18}$$

Lemma 20 and Theorem 22 of the cited Allouche--Han--Shallit paper give, respectively,

$$\operatorname{sgn}(D_m)=(-1)^{m-1}J_{m-1}, \qquad |D_m|=1+\rho(m-1). \tag{19}$$

Substitution into (18) yields

$$\mathcal H_m(xr)=-J_{m-1}\rho(m-1). \tag{20}$$

Therefore (17) becomes the exact signed formula

$$\mathcal H_k(G)=(-1)^{k+1}J_k\rho(k). \tag{21}$$

The standard Rueppel determinant evaluation and parity decomposition give

$$\begin{aligned} R_j&=\det(r_{a+b})_{0\leq a,b<j} =(-1)^{\binom j2},\\ J_{2j}&=R_jJ_j,\\ J_{2j+1}&=R_{j+1}J_j. \end{aligned} \tag{22}$$

In particular, $J_j\in\{-1,1\}$ for every $j\geq0$.

Taking $k=N-2=n-1$ in (7) and (21) now gives both signed identities in (4). The cases $n=0,1$ follow directly from the first coefficients $1,-1,1$.

Finally, replacing the linear term $-x$ by $+x$ replaces $F_{\pm}(x)$ by $F_{\pm}(-x)$. By diagonal conjugation of the Hankel matrices,

$$\mathcal H_N(F_{\pm}(-x))=\mathcal H_N(F_{\pm}(x)). \tag{23}$$

Therefore the corrected binary-run formula holds for all four independent choices of linear and denominator signs, even though the original unsigned equality is already refuted at $n=2$.