The A268411 product formula for Rueppel Hankel transforms

About 5 years old · traced to 2021

Let $h_n$ be the Hankel transform of the Rueppel sequence $r_n$ , and let $H_n$ be the Hankel transform of the once-shifted Rueppel sequence $r_{n+1}$ . A268411 product conjecture. The sequence

\frac{1+(-1)^n h_nH_n}{2}

is OEIS sequence A268411 evaluated at $n+1$ , where A268411 gives the parity of the number of runs of $1$ 's in the binary representation of its argument. This is the paper's concluding conjecture about a product of Hankel transforms; no proof or resolution is supplied.

References

Primary source

Paul Barry, “Conjectures and results on some generalized Rueppel sequences”, arXiv:2107.00442 (2021).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The conjecture remains unconfirmed: a complete proof has been posted, but no independent verification of it has appeared.

Paul Barry’s 2021 note proposes the product formula connecting the two Rueppel Hankel transforms with the binary-run parity sequence $\mathrm{A268411}$ . The paper explicitly presents this as Conjecture 23 and supplies no proof or resolution.

Known results

A related paper records the ordinary transform formula $h_n=(-1)^{\binom{n+1}{2}}$ and the shifted-transform recurrences $H_0=h_0$ , $H_{2n}=H_{n-1}h_n$ , and $H_{2n+1}=H_nh_n$ .

Posted attempt

A reader-written argument claims a complete proof: parity-block decomposition gives the shifted recurrences, and induction matches the resulting signs with the binary-run recurrence. The attempt has not been independently verified.

Current status (as of August 2026): The conjecture has an unverified complete-proof claim, but no independent verification or peer-reviewed resolution is recorded.

Sources

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Proof

Let

h_n=\det(r_{i+j})_{0\leq i,j\leq n}, \qquad H_n=\det(r_{i+j+1})_{0\leq i,j\leq n},

where $r_j=1$ exactly when $j=2^a-1$ for some $a\geq0$ , and $r_j=0$ otherwise.

The shifted Rueppel sequence is an aeration of the original sequence:

r_{2j+1}=r_j, \qquad r_{2j+2}=0.

Order the rows and columns of the matrix defining $H_n$ by parity. Because all mixed-parity entries vanish, the matrix becomes block diagonal. For $n=2m$ , its two blocks have determinants $h_m$ and $H_{m-1}$ ; for $n=2m+1$ , they have determinants $h_m$ and $H_m$ . Hence

H_0=h_0=1,

H_{2m}=H_{m-1}h_m \quad (m\geq1), \qquad H_{2m+1}=H_mh_m \quad (m\geq0). \qquad (1)

The ordinary Rueppel Hankel determinants satisfy

h_n=(-1)^{\binom{n+1}{2}}. \qquad (2)

This is also the evaluation recorded with the aeration recurrence in Proposition 4 of:

https://arxiv.org/abs/2005.04066

Put $\epsilon_m=(-1)^m$ and

s_n=(-1)^nh_nH_n.

Equation (2) gives

h_{2m}=\epsilon_m, \qquad h_{2m+1}=\epsilon_{m+1}, \qquad h_m=\epsilon_mh_{m-1}\quad(m\geq1).

Substitution into (1) now gives

s_0=1, \qquad s_{2m+1}=s_m, \qquad s_{2m+2}=\epsilon_ms_m. \qquad (3)

Let $R(u)$ denote the number of maximal runs of $1$ 's in the ordinary binary expansion of $u$ , with $R(0)=0$ . Appending a zero preserves the number of runs, while appending a one creates a new run exactly when the preceding number is even. Thus

R(2u)=R(u),

and

R(2u+1)= \begin{cases} R(u)+1,&u\text{ even},\\ R(u),&u\text{ odd}. \end{cases}

Define

q_n=(-1)^{R(n+1)+1}.

Then $q_0=1$ . Applying the preceding binary recurrences to $2m+2=2(m+1)$ and $2m+3=2(m+1)+1$ gives

q_{2m+1}=q_m, \qquad q_{2m+2}=\epsilon_mq_m. \qquad (4)

Equations (3) and (4) give the same initial value and reduce every positive index to a smaller one. Strong induction therefore yields

s_n=q_n

for every $n\geq0$ . Consequently,

\frac{1+(-1)^nh_nH_n}{2} = \frac{1+(-1)^{R(n+1)+1}}{2} = R(n+1)\pmod 2.

The right-hand side is exactly $\operatorname{A268411}(n+1)$ . This proves Conjecture 23.

Conjecture source: https://arxiv.org/abs/2107.00442