Let C_m denote the mth Catalan number, and let a,b,n,m,k be as in the stated formula. The Hankel transform of the sequence aC_n+m+bC_n+m+1 is expressed as a polynomial in a and b with coefficients T_n,k,m, where T_n,k,m=fracC_m binomm+k-2m-2binomn+k+2m-22k+2m-3∏_j=0^lfloor fracm-12 rfloor-1binom2n+2m-2j-12m-4j-5∏_j=0^m-3(2m-j-2)binom2m-22m-3∏_j=0^lfloor fracm-12 rfloor-1 binom2m-2j-12m-4j-5 ∏_j=0^m-3(n+k+2m-j-2). General Hankel-transform conjecture. The Hankel transform is given by ∑_k=0^n+1 T_n,k,m b^n-k+1a^k. The paper identifies this as the central conjecture of the note and says that it generalizes an earlier formula. The supplied text gives no evidence that the formula has been proved or disproved for arbitrary parameters.

General formula conjecture for Hankel transforms of shifted Catalan combinations

From papers

Let $C_m$ denote the $m$ th Catalan number, and let $a,b,n,m,k$ be as in the stated formula. The Hankel transform of the sequence $aC_{n+m}+bC_{n+m+1}$ is expressed as a polynomial in $a$ and $b$ with coefficients $T_{n,k,m}$ , where

T_{n,k,m}=\frac{C_m \binom{m+k-2}{m-2}\binom{n+k+2m-2}{2k+2m-3}\prod_{j=0}^{\lfloor \frac{m-1}{2} \rfloor-1}\binom{2n+2m-2j-1}{2m-4j-5}\prod_{j=0}^{m-3}(2m-j-2)}{\binom{2m-2}{2m-3}\prod_{j=0}^{\lfloor \frac{m-1}{2} \rfloor-1} \binom{2m-2j-1}{2m-4j-5} \prod_{j=0}^{m-3}(n+k+2m-j-2)}.

General Hankel-transform conjecture. The Hankel transform is given by

\sum_{k=0}^{n+1} T_{n,k,m} b^{n-k+1}a^k.

The paper identifies this as the central conjecture of the note and says that it generalizes an earlier formula. The supplied text gives no evidence that the formula has been proved or disproved for arbitrary parameters.

Progress summary

Open

The formula remains an unverified conjecture: earlier special cases are known, but no reliable source settles the arbitrary-shift version.

A note dated November 2020 proposes a closed formula for the Hankel transform of the sequence $aC_{n+m}+bC_{n+m+1}$ and identifies it as its central conjecture. The retrieved paper does not prove or disprove the formula for arbitrary parameters.

Known results

The unshifted consecutive pair $\alpha C_n+\beta C_{n+1}$ satisfies a recurrence for its Hankel transform: $h_{n+2}-(\alpha+2\beta)h_{n+1}+\beta^2h_n=0$ .
Earlier formulas cover shifted Catalan numbers and the case $m=0$ , but do not establish the general formula.

Current status (as of August 2026): The arbitrary-parameter conjecture remains open publicly, with special cases known but no retrieved verified proof, disproof, or independently corroborated correction.

Sources

Sources & referencesView supporting material

Primary source

Paul Barry, “Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers”, arXiv:2011.10827 (2020).

1 week ago

Counterexample and off-by-one correction

Write $q$ for the sequence index. Take

n=1,\qquad m=2,\qquad a=1,\qquad b=0.

The conjecture as printed concerns the sequence $C_{q+2}$ . Its first nontrivial Hankel determinant is

\det \begin{pmatrix} C_2&C_3\\ C_3&C_4 \end{pmatrix} = \det \begin{pmatrix} 2&5\\ 5&14 \end{pmatrix} =3.

For $m=2$ , however, the displayed coefficient simplifies to

T_{n,k,2}=\binom{n+k+2}{2k+1}.

When $b=0$ , only the term $k=n+1=2$ remains, giving

T_{1,2,2}=\binom55=1.

Thus the conjectured right-hand side is $1$ , whereas the stated Hankel determinant is $3$ . Therefore the conjecture is false as printed.

This is not a MathDB transcription error: Conjecture 2 in the source explicitly names $aC_{q+m}+bC_{q+m+1}$ .

The source's own coefficient tables reveal the likely correction. For $m=2$ , the row $n=1$ is $(3,4,1)$ , representing

a^2+4ab+3b^2 = \det\left( aC_{i+j+1}+bC_{i+j+2} \right)_{0\le i,j\le1}.

Thus the displayed coefficients correspond to

aC_{q+m-1}+bC_{q+m},

one shift lower than the sequence named in the conjecture.

More generally, writing

D_{n,s}(a,b)= \det\left(aC_{i+j+s}+bC_{i+j+s+1}\right)_{0\le i,j\le n},

the corrected identity is

D_{n,m-1}(a,b) = \sum_{k=0}^{n+1}T_{n,k,m}b^{\,n-k+1}a^k.

A Catalan-moment and one-factor Christoffel determinant calculation reduces this identity to the coefficients of a shifted Jacobi polynomial $P_{n+1}^{(1/2,m-3/2)}$ ; factorial simplification gives exactly the printed $T_{n,k,m}$ . This proves the corrected identity for $n\ge0$ , $m\ge2$ , and arbitrary $a,b$ .

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Samuel Schlesinger · 5 days ago