Let k be a positive integer, and let (s^k)_n denote the corresponding power-sum sequence satisfying a linear recurrence (s^k)_n=∑_j=1c_j(q)(s^k)_n-j. Recurrence and coefficient conjecture. The linear recurrence corresponding to k has order lfloor k/2 rfloor+3, and each coefficient c_j(q) is a linear polynomial in q. This conjecture summarizes the patterns observed in the computed recurrence coefficients; the supplied text gives no evidence that either assertion has been proved or disproved.

Conjecture on recurrence orders and coefficient polynomials for power sums

About 9 years old · traced to 2017

Let $k$ be a positive integer, and let $(s^k)_n$ denote the corresponding power-sum sequence satisfying a linear recurrence

(s^k)_n=\sum_{j=1}c_j(q)(s^k)_{n-j}.

Recurrence and coefficient conjecture. The linear recurrence corresponding to $k$ has order $\left\lfloor k/2\right\rfloor+3$ , and each coefficient $c_j(q)$ is a linear polynomial in $q$ . This conjecture summarizes the patterns observed in the computed recurrence coefficients; the supplied text gives no evidence that either assertion has been proved or disproved.

References

Primary source

László Németh and László Szalay, “Power sums in hyperbolic Pascal triangles”, arXiv:1703.04938 (2017).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written calculation claims the conjecture is only partly right: the predicted order fails at $k=9$ , while the linear dependence on $q$ is claimed to hold generally, but neither claim has independent verification.

The conjecture predicts a uniform recurrence order and coefficients with only linear dependence on $q$ . Németh and Szalay (2017) introduced the transfer-matrix method and computed cases through $k=11$ , which motivated these patterns.

Known results

Németh and Szalay (2017): power sums for the ${4,q}$ hyperbolic Pascal triangles were reduced to linear recurrences and computed for $2\le k\le11$ .

Posted attempt

A reader-written attempt claims a complete resolution, not merely partial progress: for $k=9$ , the minimal recurrence order is exactly $6$ , contradicting the predicted $7$ ; it also claims the coefficient polynomials are affine in $q$ for all $k$ , via a rank bound and a rank-one perturbation argument. The attempt has not been independently verified.

Current status (as of August 2026): The $k=9$ order prediction is challenged by an unverified counterexample, and the affine-coefficient assertion is supported only by an unverified posted argument; no independently confirmed resolution is recorded.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

The exact-order assertion is false, although the affine-coefficient assertion is true.

Take $k=9$ , and write $S_n=(s^9)_n$ . The defining transfer system yields, for every admissible $q\ge5$ and every $n\ge7$ ,

\begin{aligned} S_n={}&(q+62)S_{n-1}+(447q+1288)S_{n-2} +(2433q-15116)S_{n-3}\\ &+(10555-2431q)S_{n-4} +(3662-450q)S_{n-5}-450S_{n-6}. \end{aligned}

These are exactly the six coefficients printed in the source's $k=9$ row; its purported seventh coefficient is explicitly $c_7(q)=0$ . Thus the recurrence has order at most six, although the conjecture predicts

\left\lfloor\frac92\right\rfloor+3=7.

Moreover, six is the exact minimal order. At $q=5$ , the first eleven values are

2,\ 514,\ 39880,\ 4470930,\ 438529598,\ 45014063550,\ 4547625234042,\ 461974238315026,\ 46840025812751018,\ 4752325523662690738,\ 482052722208599420434.

Their Hankel determinant is

\det(S_{i+j+1})_{0\le i,j<6} =-2^{23}3^{22}5^97^7\cdot31\cdot71\ne0.

Any recurrence of order at most five would make this determinant vanish. Consequently the minimal order is exactly six, not seven.

A corrected general statement also follows directly from the source transfer matrix $M_k(q)$ . If $C_j$ denotes its $j$ -th column, then

C_j=C_{k-j}\quad(1\le j<k), \qquad 2C_{k+1}=\sum_{j=1}^{k-1}C_j.

Hence

\operatorname{rank}M_k(q)\le\left\lfloor\frac k2\right\rfloor+2,

and the homogeneous characteristic annihilator

(X-1)\det(XI-M_k(q))

gives an eventual recurrence of order at most $\lfloor k/2\rfloor+3$ . Equality need not hold, as $k=9$ shows.

Finally, homogenizing the affine transfer recurrence gives the augmented matrix

A(q)=A(0)+q(e_k+e_{k+1}) (e_0^{\mathsf T}+e_k^{\mathsf T}-2e_{k+2}^{\mathsf T}).

The $q$ -dependent perturbation has rank one, so determinant multilinearity makes $\det(XI-A(q))$ affine in $q$ . The same holds after removing its initial powers of $X$ . Thus all resulting characteristic recurrence coefficients have degree at most one in $q$ , proving the conjecture's second assertion while disproving its first.