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

At least 5 years old · documented by 2020

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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A calculation claims the printed formula is false and that a one-index-lower correction is provable, but neither claim has independent verification.

Paul Barry posed the formula as Conjecture 2 in 2020, generalizing Hankel-transform formulas for shifted Catalan numbers and consecutive pairs. The paper verifies small shifts but explicitly leaves the arbitrary-shift formula conjectural.

Known results

French (2011) derived recurrences for Hankel transforms of combinations involving up to four adjacent Catalan numbers and stated further conjectures.
Barry (2020) checked and tabulated special cases $m=0,1,2,3$ , plus explicit examples, without proving the general formula.

Posted attempt

An unverified calculation takes $n=1$ , $m=2$ , $a=1$ , and $b=0$ , obtaining determinant $3$ but conjectured value $1$ ; it therefore claims the printed statement is false. It further claims the coefficients instead give $D_{n,m-1}(a,b)$ and supplies a Catalan-moment proof for that corrected identity, but the argument has not been independently verified.

Current status (as of August 2026): The published formula remains an explicit conjecture, while a posted counterexample and claimed one-shift correction are unverified, so the original problem is not independently settled.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

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 $a,b$ .