Fix r≥ 2 and k≥ 2, and let v_r^(k)(n,t) be the sum of the rth powers of the coefficients of ∏_i=1^n(1+tx^F_i+k-1^(k)), with J_r^(k)(t,x)=∑_n≥ 0v_r^(k)(n,t)x^n. Denominator-shape conjecture. There is an integer m≥ 1, depending on r, and polynomials a_i(t), depending on r but independent of k, such that J_r^(k)(t,x)=frac1+a_2(t)x^k+a_4(t)x^2k+…+a_2m(t)x^mkD_r^(k)(t,x), where D_r^(k)(t,x)=1+a_1(t)x+a_2(t)x^k+a_3(t)x^k+1+a_4(t)x^2k+a_5(t)x^2k+1+…+a_2m(t)x^mk+a_2m+1(t)x^mk+1, with a_2m+1(t)≠ 0; possibly a_i(t)≠ 0 for every 0≤ i≤ 2m+1. Moreover, the largest index j for which a_j(1)≠ 0 is odd. This conjecture is motivated by the explicit cases for r=3 and for r=4,5,6,7 at t=1, and by the observed odd denominator degrees. The source provides empirical evidence but no proof.

Denominator-shape conjecture for generalized Fibonacci coefficient moments

About 5 years old · traced to 2021

Fix $r\geq 2$ and $k\geq 2$ , and let $v_r^{(k)}(n,t)$ be the sum of the $r$ th powers of the coefficients of

\prod_{i=1}^n\left(1+tx^{F_{i+k-1}^{(k)}}\right),

with

J_r^{(k)}(t,x)=\sum_{n\geq 0}v_r^{(k)}(n,t)x^n.

Denominator-shape conjecture. There is an integer $m\geq 1$ , depending on $r$ , and polynomials $a_i(t)$ , depending on $r$ but independent of $k$ , such that

J_r^{(k)}(t,x)=\frac{1+a_2(t)x^k+a_4(t)x^{2k}+\cdots+a_{2m}(t)x^{mk}}{D_r^{(k)}(t,x)},

where

D_r^{(k)}(t,x)=1+a_1(t)x+a_2(t)x^k+a_3(t)x^{k+1}+a_4(t)x^{2k}+a_5(t)x^{2k+1}+\cdots+a_{2m}(t)x^{mk}+a_{2m+1}(t)x^{mk+1},

with $a_{2m+1}(t)\neq 0$ ; possibly $a_i(t)\neq 0$ for every $0\leq i\leq 2m+1$ . Moreover, the largest index $j$ for which $a_j(1)\neq 0$ is odd.

This conjecture is motivated by the explicit cases for $r=3$ and for $r=4,5,6,7$ at $t=1$ , and by the observed odd denominator degrees. The source provides empirical evidence but no proof.

References

Primary source

Richard P. Stanley, “Theorems and Conjectures on Some Rational Generating Functions”, arXiv:2101.02131 (2021).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

The general denominator pattern remains unproved, while a reader-written argument claims a complete proof of a corrected version with an extra parameter, but that argument has not been independently verified and does not establish the stronger statement posed here.

The conjecture predicts a uniform rational-function denominator pattern for all moment orders $r\geq 2$ and Fibonacci orders $k\geq 2$ , together with an odd-index condition at $t=1$ . Richard Stanley formulated the relevant general conjecture in 2021.

Known results

The case $r=2$ is proved for every $k\geq 2$ (Stanley, 2021).
The explicit $r=3$ formula is conjectural in general and was verified for $k\leq 5$ by Doron Zeilberger.
At $t=1$ , explicit conjectural forms are recorded for $r=4,5,6,7$ .
Related numerator and odd-degree patterns were checked empirically through $n\leq 17$ .

Posted attempt

A reader-written argument claims a complete proof for the source’s corrected formulation, allowing coefficients $a_i(t,t^{k-1})$ , with $m=r-1$ and the final odd-index assertion. It explicitly identifies the extra parameter omitted by this problem; the argument has not been independently verified and does not prove the stated $a_i(t)$ -only version.

Current status (as of August 2026): The $r=2$ case and bounded or empirical cases are established, while the general conjecture as stated, including its odd-index assertion, remains unproved; a claimed proof of a corrected formulation is unverified.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

There is an important transcription correction: the latest primary source, Conjecture 5.6 of arXiv:2101.02131v3, permits coefficients aᵢ(t,u) with u=t^{k−1}. The problem page omits this indispensable second argument. Ekhad and Zeilberger, arXiv:2103.12855, had already identified the correction and verified bounded cases. The following proves the corrected source conjecture for every moment order r≥2 and every generalized Fibonacci order k≥2, including its final odd-index assertion.

Write z for the generating-function variable, q=zᵏ, u=t^{k−1}, and m=r−1. For 1≤s,j≤m define

W_{sj}(t)=1_{j≥s} binom(r−s,j−s)t^{j−s} +1_{j≤s} binom(s,j)t^{r−s+j},

U=diag(u,u²,…,u^m), c_s=binom(r,s)t^s, b_s=u^s t^{r−s}.

Set

N(q)=det(I−qUW),

D(q,z)=[1−(1+tʳ)z]N(q)−zq cᵀ adj(I−qUW)b.

Then the exact universal formula is

Jᵣ⁽ᵏ⁾(t,z)=N(zᵏ)/D(zᵏ,z).

In particular m=r−1 works for every k; the numerator has only powers z^{jk}, while the denominator has precisely the permitted powers z and z^{jk},z^{jk+1}, with coefficient polynomials in t and u independent of k.

Here is a direct proof. The primary source's proved free-word Lemma 5.2 yields an equal-weight two-row parser with states 0,±1,…,±k. Writing ε∈{−1,1}, its allowed difference transitions are

0 —0→ 0, 0 —ε→ ε, εj —ε→ ε(j+1) (1≤j<k), εk —0→ ε, εk —(−ε)→ 0.

Run r−1 such parsers, comparing each row to the first, and weight an r-bit column by t^{number of ones}. The useful states are exactly the all-zero state O and states whose nonzero coordinates all equal one common signed stage εj. Indeed, restricting to any two parsers leaves only (0,0),(s,0),(0,s),(s,s) on accepting paths: unequal positive stages, unequal negative stages, and opposite stages lie in a forward-invariant rejecting set. Conversely, every stated common-stage state opens, advances, and closes. There are exactly 1+k(2ʳ−2) useful states.

Quotient by row permutations, writing A_{s,j} when the high group has s rows and stage j. From O, the self-loop has weight 1+tʳ and the transition to A_{s,1} has weight c_s. Interior transitions A_{s,j}→A_{s,j+1} have weight t^s. At stage k, closing has weight t^{r−s}, and the total weight of transitions to A_{j,1} is W_{sj}(t), obtained by choosing the changing rows in the old high or low group.

Let O(z) and G_s(z) be accepted-continuation series at O and A_{s,1}. These exact transitions give

O=1+z[(1+tʳ)O+cᵀG],

G_s=zᵏt^{s(k−1)}[t^{r−s}O+∑{j=1}^m W{sj}(t)G_j].

Hence (I−qUW)G=qbO. Substitution and the adjugate identity prove O=N/D, and O is exactly the rth coefficient-moment generating function.

The final formal coefficient does not vanish: W(t)=I+O(t), so det(UW)=u^{r(r−1)/2}(1+O(t)). In the coefficient of zq^{r−1}, the adjugate contribution is u^{r(r−1)/2}O(t²), whereas the other contribution has leading term (−1)ʳu^{r(r−1)/2}. Substitution u=t^{k−1} preserves nonvanishing for every k≥2.

It remains to prove the source's delicate specialized odd-index statement. Put t=u=1, c_s=binom(r,s), and write 𝟙=(1,…,1)ᵀ. Then

D=N−zL, L(q)=2N(q)+q cᵀadj(I−qW)𝟙.

The matrix W is self-adjoint for C=diag(binom(r,1),…,binom(r,r−1)), since

binom(r,i)binom(r−i,j−i)=binom(r,j)binom(j,i).

An explicit solution of Wx=𝟙 is

x_j=(−1)^{j+1}/2 if r is even,

x_j=[(−1)^{j+1}(r−j)+1_{j=1}+1_{j=r−1}]/(2r) if r is odd.

Direct binomial summation gives cᵀx=1 when r is even and cᵀx=3/2 when r is odd. Self-adjointness and 𝟙=Wx imply that c annihilates ker W. Therefore

L(q)/N(q)=2−cᵀx+cᵀ(I−qW)^{-1}x → 1 if r is even, 1/2 if r is odd,

as q→∞. In particular deg L=deg N, even when W is singular. Thus the largest specialized nonzero denominator index is 2deg(N)+1, which is odd. This proves every clause of the corrected primary-source conjecture uniformly for all r,k≥2.