Let a_s(r,n) denote the number of two-toned tilings of a 1×(n+r+s) grid with r red squares, where the last s tiles must be white. For all s,r,n≥ 1, the conjectured identity is a_s(r,n)=2^n-r-1+s∑_j=0^r+1-sbinomr+1-sjbinomn+r-jn. This identity generalizes several previously established identities for two-toned tilings, but the source states that a proof has remained elusive.

Conjectured identity for two-toned tiling numbers

About 6 years old · traced to 2020

Let $a_s(r,n)$ denote the number of two-toned tilings of a $1\times(n+r+s)$ grid with $r$ red squares, where the last $s$ tiles must be white. For all $s,r,n\geq 1$ , the conjectured identity is

a_s(r,n)=2^{n-r-1+s}\sum_{j=0}^{r+1-s}\binom{r+1-s}{j}\binom{n+r-j}{n}.

This identity generalizes several previously established identities for two-toned tilings, but the source states that a proof has remained elusive.

References

Primary source

Robert Davis and Greg Simay, “Further Combinatorics and Applications of Two-Toned Tilings”, arXiv:2001.11089 (2020).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

The original universal formula was unproved, and a reader now claims it is false in part of the stated range and gives a complete correction, but that claim has not been independently checked.

Robert Davis and Greg Simay proposed the identity in 2020 for all $s,r,n\geq 1$ . Their paper states that a proof had remained elusive.

Known results

Davis and Simay (2020) recorded earlier identities for two-toned tilings, including a closed form in the special case $s=r$ .

Posted attempt

A reader claims the paper’s generating function implies the formula is valid only for $s\leq r+1$ and fails for every $s\geq r+2$ . The proposed smallest counterexample is $(r,s,n)=(1,3,1)$ , where the count is claimed to be $5$ while the conjectured sum is $0$ . This is presented as a complete disproof and correction, but it has not been independently verified.

Current status (as of August 2026): The universal conjecture has an unverified disproof and correction claim; independent confirmation of the generating-function argument and the corrected range remains outstanding.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Conjecture 2.2 is false as published, but admits a complete correction.

The primary source itself establishes the generating function

A_{r,s}(x) =\sum_{n\ge0}a_s(r,n)x^n =\frac{(1-x)^{r+1-s}}{(1-2x)^{r+1}}.

Put $d=r+1-s$ .

If $d\ge0$ , the finite binomial theorem gives

(1-x)^d =2^{-d}\bigl(1+(1-2x)\bigr)^d =2^{-d}\sum_{j=0}^{d}\binom dj(1-2x)^j.

Consequently

a_s(r,n) =2^{n-d}\sum_{j=0}^{d} \binom dj\binom{n+r-j}{n},

which is precisely the proposed identity. Thus the conjecture is true throughout the corrected range $s\le r+1$ .

If $d<0$ , write $h=s-r-1>0$ . The negative-binomial theorem instead gives the exact positive formula

\boxed{ a_s(r,n) = \sum_{j=0}^{n} \binom{h+j-1}{j} 2^{n-j} \binom{n-j+r}{r}>0. }

But the proposed upper summation limit is $d<0$ , so its right-hand side is an empty sum and equals zero. Hence the published claim fails for every $s\ge r+2$ .

For the smallest strictly positive example, take $r=1$ , $s=3$ , $n=1$ . Then

a_3(1,1) = [x]\frac1{(1-x)(1-2x)^2} =5,

whereas the conjectured formula gives zero. The source's own Table 3 similarly gives $a_4(2,1)=7$ , again contradicting its proposed empty sum.

Therefore the unrestricted conjecture is false, its exact valid range is $s\le r+1$ , and the omitted range $s\ge r+2$ is completely described by the positive convolution above.

Source: R. Davis and G. Simay, “Further Combinatorics and Applications of Two-Toned Tilings,” The Fibonacci Quarterly 58 (2020), 300–320, Definition 2.1 and Conjecture 2.2.