Let A_abcd and B_2-a,2-b,2-c,2-d be the paired polynomials indexed by a,b,c,d. Here a,b,c,d range over the indexing choices used to define these polynomials, and the complementary index is (2-a,2-b,2-c,2-d). Negative-coefficient conjecture. In the remaining 33 pairs of the polynomials A_abcd and B_2-a,2-b,2-c,2-d, including the case where p≠ 2 for a=b=c=d=1, one of the two polynomials will have a negative coefficient. The claim is presented as being supported by extensive testing and as an invitation to find proofs. It concerns the obstruction to obtaining nonnegative-coefficient generating polynomials in the remaining cases; no proof or resolution is given in the supplied text.

Negative-coefficient conjecture for complementary polynomial pairs

Let $A_{abcd}$ and $B_{2-a,2-b,2-c,2-d}$ be the paired polynomials indexed by $a,b,c,d$ . Here $a,b,c,d$ range over the indexing choices used to define these polynomials, and the complementary index is $(2-a,2-b,2-c,2-d)$ .

Negative-coefficient conjecture. In the remaining 33 pairs of the polynomials $A_{abcd}$ and $B_{2-a,2-b,2-c,2-d}$ , including the case where $p\neq 2$ for $a=b=c=d=1$ , one of the two polynomials will have a negative coefficient.

The claim is presented as being supported by extensive testing and as an invitation to find proofs. It concerns the obstruction to obtaining nonnegative-coefficient generating polynomials in the remaining cases; no proof or resolution is given in the supplied text.

Counterexample: infinitely many unlisted pairs

Conjecture 6.8 is false for infinitely many distinct prime triples.

Fix $p=2$ , $q=3$ , and let $r\equiv1\pmod6$ be any prime. Choose the unlisted index

(a,b,c,d)=(2,2,1,1), \qquad (2-a,2-b,2-c,2-d)=(0,0,1,1).

The two complementary source polynomials are

A=x\Phi_2^2\Phi_3^2\Phi_6^2\Phi_{2r}^2\Phi_{3r}\Phi_{6r}, \qquad B=x\Phi_r^2\Phi_{3r}\Phi_{6r}.

Cyclotomic identities give

\frac{A(x)}x =(1+x^r)^2(1+x^2+x^4)(1+x^{2r}+x^{4r}),

so every coefficient of $A$ is nonnegative. Likewise,

\frac{B(x)}x =\frac{(1+x)(1-x^r)^2(1+x^{2r}+x^{4r})} {(1-x)(1-x^6)}.

Put

g_m= \begin{cases} 2\lfloor m/6\rfloor+1+\mathbf1_{6\nmid m},&m\ge0,\\ 0,&m<0. \end{cases}

Then

[x^m]\frac{B(x)}x =\sum_{i=0}^{6}\alpha_i g_{m-ir}, \qquad (\alpha_0,\ldots,\alpha_6)=(1,-2,2,-2,2,-2,1).

Write $r=6h+1$ and $m=jr+6u+v$ , where $0\le j\le5$ , $0\le v<6$ , and $0\le6u+v<r$ . Direct substitution yields the complete coefficient table:

\begin{array}{c|rrrrrr} j\backslash v&0&1&2&3&4&5\\ \hline 0&2u+1&2u+2&2u+2&2u+2&2u+2&2u+2\\ 1&2h-2u&2h-2u-2&2h-2u-2&2h-2u-2&2h-2u-2&2h-2u-1\\ 2&2u&2u+2&2u+2&2u+2&2u+3&2u+2\\ 3&2h-2u&2h-2u-2&2h-2u-2&2h-2u-1&2h-2u-2&2h-2u-2\\ 4&2u&2u+2&2u+3&2u+2&2u+2&2u+2\\ 5&2h-2u&2h-2u-1&2h-2u-2&2h-2u-2&2h-2u-2&2h-2u-2. \end{array}

For $v=0$ , one has $u\le h$ ; for $v>0$ , one has $u\le h-1$ . Every entry is therefore nonnegative. Since $\deg(B/x)=6r-6$ , these cases exhaust all coefficients.

For the smallest instance $r=7$ ,

A(1)=36,\qquad B(1)=49,\qquad A(x)B(x)=x^2\left(\frac{1-x^{42}}{1-x}\right)^2.

Thus the omitted index gives valid 36-sided and 49-sided dice with exactly the sum frequencies of two standard 42-sided dice.

This is not an unconditional missing table row: if instead $r=6h+5$ , the same index gives

[x^{2r-4}](B/x) =g_{12h+6}-2g_{6h+1} =-1.

Consequently this unlisted family is valid precisely when $r\equiv1\pmod6$ . Dirichlet's theorem supplies infinitely many such primes, disproving Conjecture 6.8.

Negative-coefficient conjecture for complementary polynomial pairs

Progress summary

Primary source

Solutions 1

Counterexample: infinitely many unlisted pairs