Let F_n be the nth Fibonacci number, let e(n,i,k) be the exponent used to determine the sign in the source, and let d_k(r) be the Fibonomial coefficient corresponding to binomk+r-1r. For positive integers n and r, define the n× n matrix D(n,r) by D_ij=D_ij(n,r)=∑_k=0^j-1(-1)^e(n,i,k)Biggl(binomn+i+r-2iBiggr)Biggl(binomniBiggr)Biggl(binomn+k+r-2kBiggr)Biggl(binomnkBiggr)fracF_i^2∏_l=0^r-3F_i+j+lF_r∏_l=0^r-2F_i+k+l. Here the parenthesized binomial symbols are Fibonomial coefficients. Fibonomial reciprocal Hankel inverse conjecture. The matrix D(n,r) is the inverse of R_n(d_k). The surrounding discussion presents this as a proposed formula for the inverse of reciprocal Hankel matrices based on Fibonomial coefficients; the source does not establish the formula in the supplied passage.

Fibonomial reciprocal Hankel inverse formula conjecture

About 27 years old · traced to 1999

Let $F_n$ be the $n$ th Fibonacci number, let $e(n,i,k)$ be the exponent used to determine the sign in the source, and let $d_k(r)$ be the Fibonomial coefficient corresponding to $\binom{k+r-1}{r}$ . For positive integers $n$ and $r$ , define the $n\times n$ matrix $D(n,r)$ by

D_{ij}=D_{ij}(n,r)=\sum_{k=0}^{j-1}(-1)^{e(n,i,k)}\Biggl(\binom{n+i+r-2}{i}\Biggr)\Biggl(\binom{n}{i}\Biggr)\Biggl(\binom{n+k+r-2}{k}\Biggr)\Biggl(\binom{n}{k}\Biggr)\frac{F_i^2\prod_{l=0}^{r-3}F_{i+j+l}}{F_r\prod_{l=0}^{r-2}F_{i+k+l}}.

Here the parenthesized binomial symbols are Fibonomial coefficients. Fibonomial reciprocal Hankel inverse conjecture. The matrix $D(n,r)$ is the inverse of $R_n(d_k)$ . The surrounding discussion presents this as a proposed formula for the inverse of reciprocal Hankel matrices based on Fibonomial coefficients; the source does not establish the formula in the supplied passage.

References

Primary source

Thomas M. Richardson, “The Filbert Matrix”, arXiv:math/9905079 (1999).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A posted calculation claims the printed formula fails in a $1$ -by- $1$ case, but its proposed sign repair and the original conjecture remain unverified.

T. M. Richardson posed Conjecture 6.1 in 1999, asserting that the displayed matrix $D(n,r)$ inverts the reciprocal Hankel matrix built from Fibonomial coefficients.

Known results

Richardson (1999) reports computational verification for $n \le 16$ and $r \le 10$ .
The ordinary-binomial reciprocal Hankel analogue is proved; the Fibonomial statement is proposed as an analogue.
Richardson suggests that methods used for the Filbert matrix and ordinary-binomial case may prove the conjecture, but supplies no proof.
The paper records an integrality connection with the corresponding ordinary-binomial matrices.

Posted attempt

A posted calculation claims that $n=1$ , $r=3$ gives $D(1,3)=[-1]$ while the target inverse is $[1]$ , so the formula is false as printed. It proposes a column-sign correction, matching checks for $1\le n\le 8$ and $2\le r\le 10$ , but the calculation and correction have not been independently verified.

Current status (as of August 2026): The printed conjecture has an unverified claimed counterexample, while the proposed corrected formula and any general proof remain open.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Counterexample to the formula as printed

Take $n=1$ and $r=3$ . Then

d_1(3)=\binom{3}{3}_F=1,

so $R_1(d_k)=[1]$ , whose inverse is also $[1]$ .

For the proposed matrix $D(1,3)$ , set $i=j=1$ . The defining sum contains only the term $k=0$ , and the printed exponent gives

e(1,1,0) = 1(1+0+1)+\binom{1}{2}+\binom{0}{2}+1 =3.

The unsigned part of the summand is

\binom31_F\binom11_F\binom20_F\binom10_F \frac{F_1^2F_2}{F_3F_1F_2} = 2\cdot1\cdot1\cdot1\cdot\frac12 =1.

Hence

D(1,3)=[-1], \qquad D(1,3)R_1(d_k)=[-1]\ne[1].

Therefore the displayed formula for $D(n,r)$ is false as printed.

Scope and likely explanation

This refutes only the displayed formula. I checked both the 1999 arXiv version and the published 2001 Fibonacci Quarterly version; both print the same exponent and formula. Thus this is not a MathDB transcription error.

The paper nevertheless reports computational verification for $n\le16$ and $r\le10$ , which includes the example above. The most likely explanation is therefore a typographical sign discrepancy between the displayed formula and the formula used in those computations—not a failure of the intended conjecture.

Multiplying column $j$ by $(-1)^{n+j+1}$ repairs every case checked exactly for $1\le n\le8$ and $2\le r\le10$ . I do not claim an all-index proof of this corrected formula.