Conjectured identity for two-toned tiling numbers

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.