Proof of a Conjecture of Stanley About Stern’s Array

David E. Speyer Department of Mathematics 2844 East Hall 530 Church Street Ann Arbor, MI 48109-1043 USA

                   speyer@umich.edu

Abstract

Stanley, building on work of Stern, defined an array of numbers as follows: the first row of the array is ⋯ 0001000 · · ·. Each successive row is obtained by copying down the previous row and inserting, between each pair of numbers from the previous row, the sum of that pair of numbers. Let s_{n}^{r} be the sum of the r-th powers of the elements in the n-th row of the array. Stanley showed that, for each positive integer r, the sequence s_{n}^{r} obeys a homogeneous linear recurrence in n of length r/2 + O(1). Numerical evidence, however, suggested that s_{n}^{r} obeys shorter recurrences, of length r/3 + O(1). We prove Stanley’s conjecture.

1 Introduction and main results

Richard Stanley [2], following a construction of Stern [3], introduced an array of numbers defined as follows: We start with the sequence ⋯ 0001000 · · ·. We copy down this sequence and, in between each two consecutive elements of this sequence, we insert the sum of those two elements to obtain ⋯ 0000011100000 ⋯; the new elements are shown in red. We repeat this process to obtain the array shown below (with 0’s omitted).

1

1  1  1
1  1  2  1  2  1  1
1 1 2 1 3 2 3 1 3 2 3 1 2 1 1

We label the nonzero elements of the n-th row as ⟨ \begin{matrix}n \\ k \end{matrix} ⟩ for 1 ≤ k ≤ 2^{n} − 1. So we have the recurrence:

⟨\begin{matrix}n \\ 2k \end{matrix}⟩ = ⟨\begin{matrix}n − 1 \\ k \end{matrix}⟩ and ⟨\begin{matrix}n \\ 2k + 1 \end{matrix}⟩ = ⟨\begin{matrix}n − 1 \\ k \end{matrix}⟩ + ⟨\begin{matrix}n − 1 \\ k + 1 \end{matrix}⟩

with initial conditions

⟨\begin{matrix}1 \\ 1 \end{matrix}⟩ = 1 and ⟨\begin{matrix}1 \\ k \end{matrix}⟩ = 0 for k ≠ 1.

This triangular array is \underline{A337277} in the On-Line Encyclopedia of Integer Sequences [1].

For binomial coefficients, we have the easy identities ∑_{k} \binom{n}{k} = 2^{n} and ∑_{k} \binom{n}{k}^{2} = \binom{2n}{n}, but there are no simple formulas for ∑_{k} \binom{n}{k}^{r} for r ≥ 2. In contrast, Stanley showed that ∑_{k} ⟨ \begin{matrix}n \\ k \end{matrix} ⟩^{r} obeys a homogeneous linear recurrence for all r.

We repeat Stanley’s argument. For any function f(x, y) from ℝ^{2} → ℝ, put

S_{n}(f) = ∑_{k} f (⟨\begin{matrix}n \\ k \end{matrix}⟩ , ⟨\begin{matrix}n \\ k + 1 \end{matrix}⟩) .

Define (σ^{∗}f)(x, y) = f(x + y, y) and (τ^{∗}f)(x, y) = f(x, x + y). Then we have

S_{n}(f) = ∑_{i} f (⟨\begin{matrix}n \\ 2i \end{matrix}⟩ , ⟨\begin{matrix}n \\ 2i + 1 \end{matrix}⟩) + ∑_{j} f (⟨\begin{matrix}n \\ 2j + 1 \end{matrix}⟩ , ⟨\begin{matrix}n \\ 2j + 2 \end{matrix}⟩)

= ∑_{i} f (⟨\begin{matrix}n − 1 \\ i \end{matrix}⟩ , ⟨\begin{matrix}n − 1 \\ i \end{matrix}⟩ + ⟨\begin{matrix}n − 1 \\ i + 1 \end{matrix}⟩)

+ ∑_{j} f (⟨\begin{matrix}n − 1 \\ j \end{matrix}⟩ + ⟨\begin{matrix}n − 1 \\ j + 1 \end{matrix}⟩ , ⟨\begin{matrix}n − 1 \\ j + 1 \end{matrix}⟩)

= S_{n−1}(σ^{∗}f) + S_{n−1}(τ^{∗}f)
= S_{n−1}(σ^{∗}f + τ^{∗}f).

If f is a homogenous polynomial of degree r, then σ^{∗}f and τ^{∗}f are also such polynomials. Thus f ↦ σ^{∗}f + τ^{∗}f is a linear endomorphism of the vector space of degree r polynomials, given by some (r + 1) × (r + 1) matrix Φ. Since S_{n}(f) is a linear function of g, it follows that, for any degree r polynomial f, the sequence S_{n}(f) obeys the same linear recurrence as the sequence of matrix powers Φ^{n}.

Example 1. We consider cubic polynomials. A basis for the space of homogenous cubics in two variables is x^{3}, x^{2}y, xy^{2}, y^{3}, and we have

σ^{∗}(x^{3}) + τ^{∗}(x^{3}) = (x + y)^{3} + x^{3}  = 2x^{3}+3x^{2}y+3xy^{2}+ y^{3}
σ^{∗}(x^{2}y) + τ^{∗}(x^{2}y) = (x + y)^{2}y + x^{2}(x + y) = x^{3}+2x^{2}y+2xy^{2}+ y^{3}
σ^{∗}(xy^{2}) + τ^{∗}(xy^{2}) = (x + y)y^{2} + x(x + y)^{2} = x^{3}+2x^{2}y+2xy^{2}+ y^{3}
σ^{∗}(y^{3}) + τ^{∗}(y^{3}) = y^{3}  + (x + y)^{3} = x^{3}+3x^{2}y+3xy^{2}+2y^{3}.

So f ↦ σ^{∗}(f) + τ^{∗}(f) acts on this four-dimensional vector space by the matrix

Φ = [\begin{matrix}2 & 1 & 1 & 1 \\ 3 & 2 & 2 & 3 \\ 3 & 2 & 2 & 3 \\ 1 & 1 & 1 & 2 \end{matrix}] .

We say that a sequence T_{n} obeys a homogenous linear recurrence of length ℓ if there are constants a_{1}, a_{2}, …, a_{ℓ} such that T_{n} = ∑_{j=1}^{ℓ} a_{j}T_{n−j}. Since the matrix Φ in the above example is 4 × 4, the matrix powers Φ^{n} obey a homogenous linear recurrence of length 4, and thus the S_{n}(f) do likewise for any homogenous cubic polynomial f. More generally, if f is a homogenous polynomial of degree r, this argument shows that the sequence S_{n}(f) will obey a recurrence of length r + 1.

This is an elegant argument, but it falls far short of the truth. Computation shows that

(S_{n}(x^{3}))_{n≥1} = (S_{n}(y^{3}))_{n≥1} = (1, 3, 21, 147, …)
(S_{n}(x^{2}y))_{n≥1} = (S_{n}(xy^{2}))_{n≥1} = (0, 2, 14, 98, …).

We actually have the length-1 recurrence S_{n}(f) = 7S_{n−1}(f) (for n ≥ 2)!

Why do these recurrences have length 1 rather than 4? There are two reasons. The first is that the operator Φ commutes with the map f(x, y) ↦ f(y, x), and we have S_{n}(f(x, y)) = S_{n}(f(y, x)). So Φ passes to the 2-dimensional quotient vector space where we identify f(x, y) and f(y, x). Writing [f] for the class of a polynomial in this quotient, we have Φ ([x^{3}]) = 3[x^{3}]+6[x^{2}y] and Φ ([x^{2}y]) = 2[x^{3}]+4[x^{2}y], so Φ acts by the matrix Φ_{sym} = [ \begin{matrix}3 & 2 \\ 6 & 4 \end{matrix} ]. Thus S_{n}(f) obeys a recurrence of length 2. In general, this argument shows that, for f a homogenous polynomial of degree r, the sequence S_{n}(f) will obey a recurrence of length ⌈\frac{r+1}{2}⌉. (For the reader who doesn’t like quotient vector spaces, we can instead note that Φ preserves the subspace of symmetric polynomials, and that S_{n}(f(x, y)) = S_{n}(\frac{f(x,y)+f(y,x)}{2}), so we can use the matrix of Φ on this ⌈\frac{r+1}{2}⌉-dimensional subspace.)

The second reason is that the characteristic polynomial of Φ_{sym} is x^{2} − 7x, so we have Φ_{sym}^{n} = 7Φ_{sym}^{n−1} for n ≥ 2, and thus S_{n}(f) = 7S_{n−1}(f). The 0-eigenvalue makes the recurrence shorter!

Stanley conjectured that the analogous operators Φ_{sym} had repeated eigenvalues of 0 for r odd, and of ±1 for r even. The latter phenomenon makes the minimal polynomial of Φ_{sym} of lower degree than the characteristic polynomial, and thus also shortens the length of the recurrence satisfied by S_{n}(f). After introducing some general notation, we state Stanley’s conjectures, and prove them.

Fix a nonnegative integer r. Let V be the vector space of degree r homogeneous polynomials in x and y. For a 2 × 2 integer matrix γ = [ \begin{matrix}a & b \\ c & d \end{matrix}], we define

(γ^{∗}f)(x, y) = f(ax + cy, bx + dy).

We put σ = [ \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ] and τ = [ \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ], so this is consistent with our previous notation, and we note that (αβ)^{∗}(f) = β^{∗}α^{∗}f. We put Φ = σ^{∗} + τ^{∗}, a linear map V → V.

Let V_{sym} be the quotient of V by f(x, y) ≡ f(y, x). The map Φ passes to the quotient V_{sym}, and we denote this quotient map by Φ_{sym}. Write m(M, λ) for the multiplicity of λ as an eigenvalue of the matrix M.

Our main theorem involves many periodic functions of r, where r is restricted to either even or odd numbers. If r is restricted to even values, we write [a_{1}, a_{2}, …, a_{p}]_{r} for the periodic function f with f(0) = a_{1}, f(2) = a_{2}, f(4) = a_{3}, …, f(2p) = a_{1}, …, et cetera; if r is restricted to odd values, we write [a_{1}, a_{2}, …, a_{p}]_{r} for the periodic function f with f(1) = a_{1}, f(3) = a_{2}, f(5) = a_{3}, …, f(2p + 1) = a_{1}, …, et cetera.

Theorem 2. The operators Φ and Φ_{sym} are diagonalizable over ℝ. For r odd, we have

m(Φ, 0) ≥ \frac{r}{3} + [−\frac{1}{3}, 1, \frac{1}{3}]_{r} m(Φ_{sym}, 0) ≥ \frac{r}{6} + [−\frac{1}{6}, \frac{1}{2}, \frac{1}{6}]_{r} .

For r even, we have

m(Φ, 1) ≥ \frac{r}{6} + [−1, \frac{2}{3}, \frac{1}{3}, 0, −\frac{1}{3}, \frac{4}{3}]_{r}  m(Φ_{sym}, 1) ≥ \frac{r}{12} + [−1, −\frac{1}{6}, −\frac{1}{3}, −\frac{1}{2}, −\frac{2}{3}, \frac{1}{6}]_{r}
m(Φ, −1) ≥ \frac{r}{6} + [0, −\frac{1}{3}, \frac{4}{3}, −1, \frac{2}{3}, \frac{1}{3}]_{r} m(Φ_{sym}, −1) ≥ \frac{r}{12} + [0, −\frac{1}{6}, \frac{2}{3}, −\frac{1}{2}, \frac{1}{3}, \frac{1}{6}]_{r}
m(Φ, 1) + m(Φ, −1) ≥ \frac{r}{3} + [−1, \frac{1}{3}, \frac{5}{3}]_{r} m(Φ_{sym}, 1) + m(Φ_{sym}, −1) ≥ \frac{r}{6} + [−1, −\frac{1}{3}, \frac{1}{3}]_{r} .

The parts of this theorem related to Φ_{sym} were conjectured by Stanley, except that Stanley only conjectured diagonalizability over ℂ. Stanley did not consider Φ, but we think it is natural to consider this operator both because it is easier to compute with and because it would occur naturally if one considered the Stern recurrence with asymmetric initial conditions. Computations show that the lower bounds are equalities for all 0 < r ≤ 100.

Theorem 2 has the following implications for shortening the recurrence obeyed by S_{n}(f):

Corollary 3. For r odd, the sequence S_{n}(f) obeys a homogenous linear recurrence of length ≤ \frac{r}{3} + [\frac{2}{3}, 0, \frac{1}{3}]_{r} for n large. For r even, the sequence S_{n}(f) obeys a homogenous linear recurrence of length ≤ \frac{r}{3} +[4, \frac{10}{3}, \frac{8}{3}]_{r} for n large. This can be shortened to length ≤ \frac{r}{3} +[2, \frac{4}{3}, \frac{2}{3}]_{r} if we allow recurrences of the form S_{n}(f) = ∑_{j=1}^{ℓ} a_{j}S_{n−j}(f) + b + c(−1)^{n}.

2 Diagonalizability of Φ and Φ_{sym}

In this section, we will show that Φ on V is diagonalizable over ℝ. Since Φ_{sym} is the action of Φ on the quotient V_{sym}, this also establishes diagonalizability of Φ_{sym}. Working explicitly in the basis x^{b}y^{r−b} for V, we have

σ^{∗}(x^{b}y^{r−b}) = ∑_{a} \binom{b}{a}x^{a}y^{r−a} and τ^{∗}(x^{b}y^{r−b}) = ∑_{a} \binom{r − b}{r − a}x^{a}y^{r−a}

so, in this basis, the entries of the matrix Φ are given by

Φ_{ab} = \binom{b}{a} + \binom{r − b}{r − a}.

Example 4. Returning to the example of cubic polynomials, we have

Φ = [\begin{matrix}2 & 1 & 1 & 1 \\ 3 & 2 & 2 & 3 \\ 3 & 2 & 2 & 3 \\ 1 & 1 & 1 & 2 \end{matrix}] = [\begin{matrix}1 \\ 3 & 1 \\ 3 & 2 & 1 \\ 1 & 1 & 1 & 1 \end{matrix}] + [\begin{matrix}1 & 1 & 1 & 1 \\ & 1 & 2 & 3 \\ & & 1 & 3 \\ & & & 1 \end{matrix}] .

Conjugate Φ by the diagonal matrix D with entries D_{kk} = \sqrt{k!(r − k)!}. We have

(DΦD^{−1})_{ab} = \sqrt{\frac{a!(r−a)!}{b!(r−b)!}} · \begin{cases}\binom{b}{a}, & for a < b; \\ \binom{r−b}{r−a}, & for a > b; \end{cases}
= \frac{1}{|a−b|!} \sqrt{\frac{max(a,b)! max(r−a,r−b)!}{min(a,b)! min(r−a,r−b)!}}.

The last expression is symmetric in a and b, so DΦD^{−1} is a symmetric matrix and thus diagonalizable over ℝ. We conclude that Φ is likewise diagonalizable over ℝ.

3 The case of r odd

In this section and the next, we will want to discuss complex eigenvalues of Φ, so we require V to be the vector space of degree r polynomials with complex coefficients.

Throughout this section, we take r to be odd. Set

ρ = στ^{−1} = [\begin{matrix}1 & −1 \\ 1 & 0 \end{matrix}] .

Let ω be a primitive 6-th root of unity. Then ρ acts on ℂ^{2} with eigenvalues ω and ω^{−1}. Thus ρ^{∗} acts on V with eigenvalues ω^{a}(ω^{−1})^{r−a} for 0 ≤ a ≤ r for 0 ≤ a ≤ r; the corresponding eigenvectors are (x + ωy)^{a}(ωx + y)^{r−a}. In particular, the (−1)-eigenspace of ρ^{∗} on V corresponds to those values of a for which a ≡ r − a + 3 (mod 6) and has dimension equal to the number of such a, which is \frac{r}{3} + [−\frac{1}{3}, 1, \frac{1}{3}]_{r}.

Let W be this (−1)-eigenspace of ρ^{∗}. For g ∈ W, we have ρ^{∗}g = −g, so σ^{∗}g = −τ^{∗}g. Then Φ(g) = σ^{∗}g + τ^{∗}g = 0. So Φ acts by 0 on W, and thus m(Φ, 0) ≥ \frac{r}{3} + [−\frac{1}{3}, 1, \frac{1}{3}]_{r}

We now consider the corresponding computation for Φ_{sym}. Let W_{sym} be the (−1)- eigenspace of Φ_{sym} on V_{sym}. In Φ_{sym}, the eigenvectors (x + ωy)^{a}(ωx + y)^{r−a} and (x + ωy)^{r−a}(ωx + y)^{a} are identified, and no other relations are imposed. So the dimension of W_{sym} is the number of a for which a ≡ r − a + 3 (mod 6), modulo the relation that a and r −a are equivalent; the number of such a is \frac{r}{6} +[−\frac{1}{6}, \frac{1}{2}, \frac{1}{6}]_{r}. We have proved the multiplicity bound for m(Φ_{sym}, 0).

4 The case of r even

Throughout this section, we take r to be even. We continue to use the notation ρ = στ^{−1} = [ \begin{matrix}1 & −1 \\ 1 & 0 \end{matrix} ] and also introduce

ι = στ^{−1}σ = [\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}] .

The eigenvalues of ρ on ℂ^{2} are the primitive sixth roots of unity and the eigenvalues of ι are ±i. Thus, the eigenvalues of ρ^{∗} on V are cube roots of unity (not necessarily primitive) and the eigenvalues of ι^{∗} are ±1. We deduce that, on V, we have (ρ^{∗})^{3} = (ι^{∗})^{2} = Id.

Let X be the subspace of V where (ρ^{∗})^{2} + (ρ^{∗}) + 1 = 0, and let Y^{+} and Y^{−} be the subspaces of V where ι^{∗} has eigenvalues 1 and −1 respectively. We have

dim X = \frac{2r}{3} + [0, \frac{2}{3}, \frac{4}{3}]_{r}  dim Y^{+} = \frac{r}{2} + [1, 0]_{r}  dim Y^{−} = \frac{r}{2} + [0, 1]_{r}.

We deduce that

dim(X ∩ Y^{+}) ≥ \frac{2r}{3} + [0, \frac{2}{3}, \frac{4}{3}]_{r} + \frac{r}{2} + [1, 0]_{r} − (r + 1) = \frac{r}{6} + [0, −\frac{1}{3}, \frac{4}{3}, −1, \frac{2}{3}, \frac{1}{3}]_{r}
dim(X ∩ Y^{−}) ≥ \frac{2r}{3} + [0, \frac{2}{3}, \frac{4}{3}]_{r} + \frac{r}{2} + [0, 1]_{r} − (r + 1) = \frac{r}{6} + [−1, \frac{2}{3}, \frac{1}{3}, 0, −\frac{1}{3}, \frac{4}{3}]_{r}.

Adding these formulas,

dim(X ∩ Y^{+}) + dim(X ∩ Y^{−}) ≥ \frac{r}{3} + [−1, \frac{1}{3}, \frac{5}{3}]_{r}.

For f in X, we have

Φ(f) = σ^{∗}(f) + τ^{∗}(f) = τ^{∗} (ρ^{∗} + 1) (f) = −τ^{∗}(ρ^{∗})^{2}(f) = −σ^{∗}(τ^{−1})^{∗}σ^{∗}(f) = −ι^{∗}(f).

So, if f ∈ X ∩ Y^{±}, we have Φ(f) = ∓f. This proves the claimed bounds for m(Φ, ±1).

We now repeat the analysis for Φ_{sym}. We define X_{sym} and Y_{sym}^{±} analogously. A basis for X is those polynomials (x + ωy)^{a}(ωx + y)^{r−a} for which 2a ≢ r (mod 3). In X_{sym} the polynomials (x + ωy)^{a}(ωx + y)^{r−a} and (x + ωy)^{r−a}(ωx + y)^{r} are identified, which cuts the dimension exactly in half. (Note that (x + ωy)^{r/2}(ωx + y)^{r/2} has eigenvalue 1, so it doesn’t contribute to X_{sym}.) So

dim X_{sym} = \frac{r}{3} + [0, \frac{1}{3}, \frac{2}{3}]_{r}.

Similar but simpler analysis shows that

dim Y_{sym}^{+} = \frac{r}{4} + [1, \frac{1}{2}]_{r} dim Y_{sym}^{−} = \frac{r}{4} + [0, \frac{1}{2}]_{r}.

We have

dim(X_{sym} ∩ Y_{sym}^{+}) ≥ \frac{r}{3} + [0, \frac{1}{3}, \frac{2}{3}]_{r} + \frac{r}{4} + [1, \frac{1}{2}]_{r} − (\frac{r}{2} + 1) = \frac{r}{12} + [0, −\frac{1}{6}, \frac{2}{3}, −\frac{1}{2}, \frac{1}{3}, \frac{1}{6}]_{r}
dim(X_{sym} ∩ Y_{sym}^{−}) ≥ \frac{r}{3} + [0, \frac{1}{3}, \frac{2}{3}]_{r} + \frac{r}{4} + [0, \frac{1}{2}]_{r} − (\frac{r}{2} + 1) = \frac{r}{12} + [−1, −\frac{1}{6}, −\frac{1}{3}, −\frac{1}{2}, −\frac{2}{3}, \frac{1}{6}]_{r}

Adding these,

m(Φ_{sym}, 1) + m(Φ_{sym}, −1) ≥ \frac{r}{6} + [−1, −\frac{1}{3}, \frac{1}{3}]_{r} .

We have now deduced all claims about m(Φ_{sym}, ±1).

5 Consequences for the length of recurrences

First, we consider the case of r odd. The matrix Φ_{sym} is square of size \frac{r+1}{2}; let the characteristic polynomial of Φ_{sym} be x^{(r+1)/2} − ∑_{j=1}^{(r+1)/2} a_{j}x^{(r+1)/2−j}. So we have

Φ_{sym}^{n} = ∑_{j=1}^{(r+1)/2} a_{j}Φ_{sym}^{n−j}

. But the last m(Φ_{sym}, 0) terms of the sum are 0, so the sum actually only runs up to j = \frac{r+1}{2} − m(Φ_{sym}, 0) ≥ \frac{r}{3} + [\frac{2}{3}, 0, \frac{1}{3}]_{r}. For any degree r polynomial f, the sequence S_{n}(f) is a linear function of Φ_{sym}^{n}, so S_{n}(f) obeys a homogenous linear recurrence of the same length.

We now consider the case of r even. Write the characteristic polynomial of Φ_{sym} in the form g(x)(x − 1)^{m(Φ_{sym},1)}(x + 1)^{m(Φ_{sym},1)}. Then Φ_{sym} obeys the polynomial g(x)(x − 1)(x + 1) with degree \frac{r}{2} + 1 − (m(Φ_{sym}, 1) + m(Φ_{sym}, −1) − 2) = \frac{r}{3} + [4, \frac{10}{3}, \frac{8}{3}]_{r}. Hence Φ_{sym}^{n} obeys a homogenous linear recurrence of this length, and S_{n}(f) does as well.

Moreover, we can change bases to put Φ_{sym} into block form [ \begin{matrix}M & 0 & 0 \\ 0 & Id & 0 \\ 0 & 0 & −Id \end{matrix} ], so g(M) = 0

and let g(x) = ∑ g_{j}x^{deg(g)−j}. Then ∑ g_{j}Φ_{sym}^{n−j} will have block form [ \begin{matrix}0 & 0 & 0 \\ 0 & pId & 0 \\ 0 & 0 & q(−1)^{n} \end{matrix} ] and hence ∑ g_{j}S_{n−j}(f) will be of the form b + c(−1)^{n} for some constants b and c. So S_{n}(f) obeys a recurrence of the form S_{n}(f) = ∑_{j=1}^{ℓ} a_{j}S_{n−j}(f) + b + c(−1)^{n} for ℓ = deg(g).

6 Acknowledgments

The author was partially supported by NSF grant DMS-1600223. The author learned of this conjecture from Richard Stanley’s talk at “Combinatorics and Beyond”, a 2018 conference in honor of Sergey Fomin, and thanks Professor Stanley for bringing this delightful construction to his attention. Thanks to the referee for their careful reading.

References

[1] N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, online resource, available at https://oeis.org, 2022.

[2] R. Stanley, Some linear recurrences motivated by Stern’s diatomic array, Amer. Math. Monthly 127 (2020), 99–111.

[3] M. Stern, Ueber eine zahlentheoretische Funktion, J. Reine Angew. Math. 55 (1858), 193–220.

2020 Mathematics Subject Classification: Primary 05A99. Keywords: Stern sequence, Stern’s diatomic array.

(Concerned with sequence \underline{A337277}.)

Received March 15 2022; revised versions received June 19 2022; July 6 2022. Published in Journal of Integer Sequences, July 18 2022.

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