The Stieltjes-parameter formula for generalized Rueppel sequences

Proof of the generalized Rueppel Stieltjes-parameter conjecture

Let $b,c\neq0$ belong to a field of characteristic zero, and define

$$M(x)=r_{b,c}(x) =1+cx+b\sum_{j\geq2}x^{2^j-1}. \tag{1}$$

Write its Stieltjes continued fraction as

$$M(x) =\cfrac1{1-\cfrac{s_0x}{1-\cfrac{s_1x}{1-\cfrac{s_2x}{\ddots}}}}. \tag{2}$$

Let $P(k)\in\{0,1\}$ be the regular paperfolding sequence, characterized by

$$P(2k)=\frac{1+(-1)^k}{2}, \qquad P(2k+1)=P(k). \tag{3}$$

We prove, for every $k\geq0$,

$$\boxed{ \begin{aligned} s_{8k}=s_{8k+5}&=c,\\ s_{8k+1}=s_{8k+4}&=-c,\\ s_{8k+3}=-s_{8k+2}&=(-1)^k\frac{b}{c^2},\\ s_{8k+7}=-s_{8k+6}&=\frac{2P(k)-1}{b}. \end{aligned}} \tag{4}$$

This proves Conjecture 18 of Paul Barry, Conjectures and results on some generalized Rueppel sequences uniformly for all nonzero parameters $b,c$. In particular, all the Hankel minors required to define the continued fraction are nonzero.

Ordinary and shifted Rueppel determinants

Let

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

and put

$$R_m=\det(r_{i+j})_{0\leq i,j<m}, \qquad J_m=\det(r_{i+j+1})_{0\leq i,j<m}, \qquad R_0=J_0=1. \tag{5}$$

The ordinary Rueppel determinant evaluation we will derive is

$$R_m=(-1)^{\binom m2}. \tag{6}$$

Furthermore,

$$r_{2j}=\begin{cases} 1,&j=0,\\ 0,&j\geq1, \end{cases} \qquad r_{2j+1}=r_j. \tag{7}$$

Separating even and odd row and column indices in the ordinary and once-shifted Rueppel Hankel matrices gives

$$\begin{aligned} R_{2t}&=(-1)^tR_t^2, &R_{2t+1}&=(-1)^tJ_t^2,\\ J_{2t}&=R_tJ_t, &J_{2t+1}&=R_{t+1}J_t. \end{aligned} \tag{8}$$

Starting from $R_0=J_0=1$, simultaneous induction in (8) proves (6) and shows that every $J_t$ equals $1$ or $-1$.

For the twice-shifted determinants

$$L_t=\det(r_{i+j+2})_{0\leq i,j<t}, \tag{9}$$

both same-parity blocks vanish. The off-diagonal blocks are the once-shifted Rueppel matrices, so

$$L_{2u}=(-1)^uJ_u^2=(-1)^u, \qquad L_{2u+1}=0. \tag{10}$$

Exact moment determinants for both parameters

Let $\mu_j=[x^j]M(x)$. Equation (1) gives

$$\mu_{2j}=\begin{cases} 1,&j=0,\\ 0,&j\geq1, \end{cases} \qquad \mu_{2j+1}=w_j, \tag{11}$$

where

$$w_j=br_j+(c-b)\mathbf1_{\{j=0\}}. \tag{12}$$

Define

$$W_t=\det(w_{i+j})_{0\leq i,j<t}, \qquad W_0=1. \tag{13}$$

The matrix in (13) differs from $b(r_{i+j})$ only at its upper-left entry. Consequently,

$$W_t=b^{t-1}\left(bR_t+(c-b)L_{t-1}\right) \qquad(t\geq1). \tag{14}$$

When $t$ is even, (10) gives $L_{t-1}=0$. When $t=2u+1$ is odd, (6) and (10) give

$$L_{t-1}=(-1)^u=R_t.$$

Therefore

$$\boxed{ W_t=R_t\,b^{\,t-(t\bmod2)}c^{\,t\bmod2}. } \tag{15}$$

Now define the ordinary and once-shifted moment determinants

$$\Delta_N=\det(\mu_{i+j})_{0\leq i,j<N}, \qquad \Gamma_N=\det(\mu_{i+j+1})_{0\leq i,j<N}, \qquad \Delta_0=\Gamma_0=1. \tag{16}$$

For $\Delta_{2m}$, the two parity classes have equal size, and their off-diagonal block is $(w_{i+j})$. For $\Delta_{2m+1}$, expansion along the unique nonzero same-parity entry leaves two blocks $(w_{i+j+1})$, whose determinants are $b^mJ_m$. Thus

$$\boxed{ \Delta_{2m}=(-1)^mW_m^2, \qquad \Delta_{2m+1}=(-1)^mb^{2m}. } \tag{17}$$

For $\Gamma_N$, the mixed-parity blocks vanish. Its two diagonal blocks are $(w_{i+j})$ and $(w_{i+j+1})$. Hence

$$\boxed{ \Gamma_N =W_{\lceil N/2\rceil}\, b^{\lfloor N/2\rfloor}J_{\lfloor N/2\rfloor}. } \tag{18}$$

Since $b,c\neq0$, (15), (17), and (18) show that every $\Delta_N$ and $\Gamma_N$ is nonzero.

Recovery of all Stieltjes coefficients

The standard Stieltjes determinant formulas are

$$\begin{aligned} s_{2n} &=\frac{\Gamma_{n+1}\Delta_n} {\Gamma_n\Delta_{n+1}},\\ s_{2n+1} &=\frac{\Delta_{n+2}\Gamma_n} {\Delta_{n+1}\Gamma_{n+1}}. \end{aligned} \tag{19}$$

From (15)--(18), the ordinary determinants simplify to

$$\begin{aligned} \Delta_{4k}&=b^{4k},& \Delta_{4k+1}&=b^{4k},\\ \Delta_{4k+2}&=-b^{4k}c^2,& \Delta_{4k+3}&=-b^{4k+2}. \end{aligned} \tag{20}$$

Write

$$E_k=(-1)^kR_kJ_k. \tag{21}$$

Using (6) and (8), the shifted determinants in (18) become

$$\begin{aligned} \Gamma_{4k}&=E_kb^{4k},\\ \Gamma_{4k+1}&=E_kcb^{4k},\\ \Gamma_{4k+2}&=(-1)^kE_kcb^{4k+1},\\ \Gamma_{4k+3}&=(-1)^{k+1}E_kb^{4k+3}. \end{aligned} \tag{22}$$

Substitution of (20) and (22) into (19) gives

$$\begin{array}{c|c|c} n&s_{2n}&s_{2n+1}\\ \hline 4k&c&-c\\ 4k+1&(-1)^{k+1}b/c^2&(-1)^kb/c^2\\ 4k+2&-c&c\\ 4k+3&-J_kJ_{k+1}/b&J_kJ_{k+1}/b. \end{array} \tag{23}$$

It remains to identify the sign in the final row. Set

$$\varepsilon_k=J_kJ_{k+1}. \tag{24}$$

Equations (6) and (8) give

$$\begin{aligned} \varepsilon_{2t} &=R_tR_{t+1}=(-1)^t,\\ \varepsilon_{2t+1} &=J_tJ_{t+1}=\varepsilon_t. \end{aligned} \tag{25}$$

Comparing with the defining paperfolding recurrences (3), and using $\varepsilon_0=1=2P(0)-1$, we obtain

$$J_kJ_{k+1}=\varepsilon_k=2P(k)-1 \qquad(k\geq0). \tag{26}$$

Substitution into (23) gives exactly the four asserted families in (4), proving the complete two-parameter Stieltjes-continued-fraction formula.