The p-Frobenius Number for the Triple of the Generalized Star Numbers

Preprint

Article

The p-Frobenius Number for the Triple of the Generalized Star Numbers

Altmetrics

Downloads

103

Views

Comments

A peer-reviewed article of this preprint also exists.

This version is not peer-reviewed

Submitted:

26 July 2024

Posted:

29 July 2024

You are already at the latest version

Alerts

Abstract

In this paper, we give closed-form expressions of the $p$-Frobenius number for the triple of the generalized star numbers $a n(n-1)+1$ for an integer $a\ge 4$. When $a=6$, it is reduced to the famous star number. For the set of given positive integers $\{a_1,a_2,\dots,a_k\}$, the $p$-Frobenius number is the largest integer $N$ whose number of nonnegative integer representations $N=a_1 x_1+a_2 x_2+\dots+a_k x_k$ is at most $p$. When $p=0$, the $0$-Frobenius number is the classical Frobenius number, which is the central topic of the famous linear Diophantine problem of Frobenius.

Keywords:

Subject: Computer Science and Mathematics - Mathematics

1. Introduction

For an integer

a \geq 4

, define the generalized star numbers

S_{a, n}

S_{a, n} = a n (n - 1) + 1 (n \geq 0) .

(1)

When

a = 6

S_{n} = S_{6, n}

are the ordinary star numbers, which are often known as the centered 12-gonal numbers or centered dodecagonal numbers. A star number is a centered figurate number that represents a centered hexagram, such as the one that Chinese checkers is played on. As can be seen in Figure 1, the star numbers show a nice symmetry. The star numbers are used for a new set of vector-valued Teichmüller modular forms, defined on the Teichmüller space, strictly related to the Mumford forms, which are holomorphic global sections of the vector bundle [1]. The first star numbers are given by

\begin{matrix} {S_{n}}_{n \geq 0} = 1, 13, 37, 73, 121, 181, 253, 337, 433, 541, 661, \\ 793, 937, 1093, 1261, 1441, 1633, 1837, 2053, \dots \end{matrix}

([2, A.131],[3, A003154]). Some well-known formulas include:

\sum_{n = 1}^{\infty} \frac{1}{S_{n}} = \frac{π tan (π / 2 \sqrt{3})}{2 \sqrt{3}}, \sum_{n = 0}^{\infty} \frac{S_{n}}{n!} = 7 e and \sum_{n = 0}^{\infty} \frac{S_{n}}{2^{n}} = 26 .

Geometrically, the n-th generalized star number

S_{a, n}

is made up of a central point and

2 a

copies of the (

n - 1

)-th triangular number, making it numerically equal to the n-th centered (

2 a

)-gonal number, but differently arranged. Therefore,

S_{a, n}

are often called the centered

2 a

-gonal numbers. The generalized star numbers satisfy the linear recurrence equation

S_{a, n} = S_{a, n - 1} + 2 a (n - 1) .

When

a = 4, 5, 7, 8, 9, 10, 11, 12

, the centered

2 a

-gonal numbers are listed in [3, A016754, A062786, A069127, A069129, A069131, A069133, A069173, A069190], respectively.

The p-numerical semigroup

S_{p} (A)

from the set of the positive integers

{a_{1}, a_{2}, \dots, a_{k}}

(

k \geq 2

) is defined as the set of integers whose nonnegative integral linear combinations of given positive integers

a_{1}, a_{2}, \dots, a_{k}

are expressed in more than p ways [4]. For some backgrounds on the number of representations, see, e.g., [5,6,7,8]. For a set of nonnegative integers

N_{0}

, the set

N_{0} ∖ S_{p}

is finite if and only if

gcd (a_{1}, a_{2}, \dots, a_{κ}) = 1

. Then there exists the largest integer

g_{p} (A) : = g (S_{p})

N_{0} ∖ S_{p}

, which is called the p-Frobenius number. The cardinality of

N_{0} ∖ S_{p}

is called the p-genus (or the p-Sylvester number) and is denoted by

n_{p} (A) : = n (S_{p})

. This kind of concept is a generalization of the famous Diophantine problem of Frobenius ([9,10]) since

p = 0

is the case when the original Frobenius number

g (A) = g_{0} (A)

and the genus

n (A) = n_{0} (A)

When

k = 2

, there exists the explicit closed formula of the p-Frobenius number for any non-negative integer p. However, for

k = 3

, the p-Frobenius number cannot be given by any set of closed formulas, which can be reduced to a finite set of certain polynomials ([11]). Since it is very difficult to give a closed explicit formula for any general sequence for three or more variables, many researchers have tried to find the Frobenius number for special cases (see, e.g., [12,13,14]). Though it is even more difficult when

p > 0

(see, e.g., [15,16,17,18]), in [19], the p-Frobenius numbers of the consecutive three triangular numbers are studied.

In this paper, the closed-form expressions of the p-Frobenius numbers of the consecutive three star numbers

{S_{a, n}, S_{a, n + 1}, S_{a, n + 2}}

(

a \geq 4

and

n \geq 2

) are shown. We also give the explicit formula for their p-Sylvester number.

2. Basic Properties of the Generalized Star Numbers

In this section, we shall show some basic formulas for the generalized star numbers.

Proposition 1.

\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{S_{a, n}} = \frac{π tan (π \sqrt{(a - 4) / a} / 2)}{\sqrt{a (a - 4)}} (a \geq 5), \sum_{n = 0}^{\infty} \frac{S_{a, n}}{n!} = (a + 1) e, \\ \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} S_{a, n}}{n!} = \frac{a + 1}{e}, \sum_{n = 0}^{\infty} \frac{S_{a, n}}{b^{n}} = \frac{2 a b}{{(b - 1)}^{3}} + \frac{b}{b - 1} (b > 1) . \end{matrix}

Proof.

The second and third identities are trivial from the definition in (1). For the first identity, by referring to [20, Ch.25], put

g (z) : = \frac{1}{a z (z - 1) + 1} = \frac{1}{a (z - α) (z - β)},

where

α = \frac{a + \sqrt{a^{2} - 4 a}}{2 a} and β = \frac{a - \sqrt{a^{2} - 4 a}}{2 a} .

For

h (z) = π g (z) cot π z

\begin{matrix} {Res}_{z = α} h & = lim_{z \to α} (z - α) π g (z) cot π z = lim_{z \to α} \frac{π cot π z}{a (z - β)} = \frac{π cot π α}{a (α - β)} \\ = \frac{π}{\sqrt{a^{2} - 4 a}} cot (\frac{a + \sqrt{a^{2} - 4 a}}{2 a} π) \\ = - \frac{π}{\sqrt{a^{2} - 4 a}} tan (\frac{\sqrt{a^{2} - 4 a}}{2 a} π), \\ {Res}_{z = β} h & = lim_{z \to β} (z - β) π g (z) cot π z = lim_{z \to β} \frac{π cot π z}{a (z - α)} = \frac{π cot π β}{a (β - α)} \\ = \frac{- π}{\sqrt{a^{2} - 4 a}} cot (\frac{a - \sqrt{a^{2} - 4 a}}{2 a} π) \\ = - \frac{π}{\sqrt{a^{2} - 4 a}} tan (\frac{\sqrt{a^{2} - 4 a}}{2 a} π) . \end{matrix}

Hence, we get

\begin{matrix} \sum_{n = - \infty}^{\infty} \frac{1}{a n (n - 1) + 1} & = \sum_{n = - \infty}^{\infty} g (n) = - ({Res}_{z = α} h + {Res}_{z = β} h) \\ = \frac{2 π}{\sqrt{a^{2} - 4 a}} tan (\frac{\sqrt{a^{2} - 4 a}}{2 a} π) . \end{matrix}

Since

\begin{matrix} \sum_{n = - \infty}^{\infty} \frac{1}{a n (n - 1) + 1} & = \sum_{n = 1}^{\infty} \frac{1}{a n (n - 1) + 1} + \sum_{n = 0}^{\infty} \frac{1}{a (- n) (- n - 1) + 1} \\ = 2 \sum_{n = 1}^{\infty} \frac{1}{a n (n - 1) + 1}, \end{matrix}

we obtain the first identity.

For the fourth identity, consider the function

f_{b} (x) : = \sum_{n = 0}^{\infty} {(\frac{x}{b})}^{n} = \frac{b}{b - x} (b > 1) .

Since

f_{b}^{''} (x) = \sum_{n = 0}^{\infty} \frac{n (n - 1)}{b^{n}} x^{n - 2} = \frac{2 b}{{(b - x)}^{3}},

we have

\begin{matrix} \sum_{n = 0}^{\infty} \frac{S_{a, n}}{b^{n}} & = a \sum_{n = 0}^{\infty} \frac{n (n - 1)}{b^{n}} + \sum_{n = 0}^{\infty} \frac{1}{b^{n}} \\ = \frac{2 a b}{{(b - 1)}^{3}} + \frac{b}{b - 1} \end{matrix}

□

3. Apéry Set

We introduce the Apéry set [21] to obtain the formulas in this paper.

Let p be a nonnegative integer. For a set of positive integers

A = {a_{1}, a_{2}, \dots, a_{k}}

with

gcd (A) = 1

and

a_{1} = min (A)

we denote by

{Ap}_{p} (A) = Ap (a_{1}, a_{2}, \dots, a_{k}) = {m_{0}^{(p)}, m_{1}^{(p)}, \dots, m_{a_{1} - 1}^{(p)}},

the Apéry set of A, where each positive integer

m_{i}^{(p)}

(0 \leq i \leq a_{1} - 1)

satisfies the conditions:

(i) m_{i}^{(p)} \equiv i (mod a_{1}), (ii) m_{i}^{(p)} \in S_{p} (A), (iii) m_{i}^{(p)} - a_{1} \notin S_{p} (A)

Note that

m_{0}

is defined to be 0.

It follows that for each p,

{Ap}_{p} (A) \equiv {0, 1, \dots, a_{1} - 1} (mod a_{1}) .

One of the convenient formulas to obtain the p-Frobenius number is via the elements in the corresponding p-Apéry set ([22]).

Lemma 1.

Let

gcd (a_{1}, \dots, a_{k}) = 1

with

a_{1} = min {a_{1}, \dots, a_{k}}

. Then we have

\begin{matrix} g_{p} (a_{1}, \dots, a_{k}) & = max_{0 \leq j \leq a_{1} - 1} m_{j}^{(p)} - a_{1}, \end{matrix}

(2)

\begin{matrix} n_{p} (a_{1}, \dots, a_{k}) & = \frac{1}{a_{1}} \sum_{j = 0}^{a_{1} - 1} m_{j}^{(p)} - \frac{a_{1} - 1}{2} . \end{matrix}

(3)

Remark 1.

When

p = 0

, the formulas (2) and (3) are essentially due to Brauer and Shockley [23], and Selmer [24], respectively. More general formulas, including the p-power sum and the p-weighted sum, can be seen in [22,25].

4. Main Result

It has been determined that the p-Frobenius number of the three consecutive generalized star numbers can be given as follows. The general result for a non-negative p is based upon the result for

p = 0

, which is stated as follows.

Theorem 1.

When a is even with

a \geq 4

and

n \geq 2

, we have

\begin{matrix} g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + (\frac{a}{2} - l) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)} (l \geq 2); \\ (2 l n - \frac{3 a}{2} + l - 3) S_{a, n + 1} + (\frac{a}{2} - l + 1) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)} (l \geq 2); \\ (2 n) S_{a, n + 1} + (\frac{a}{2} - 1) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}; \\ (2 n - \frac{3 a}{2} - 2) S_{a, n + 1} + \frac{a}{2} n S_{a, n + 2} - S_{a, n} \\ if n \geq \frac{3 a}{2} + 1 . \end{matrix} \end{matrix}

When a is odd with

a \geq 5

and

n \geq 2

, we have

\begin{matrix} g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + \frac{a - 2 l + 1}{2} n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 7}{2 (2 l - 1)} \leq n \leq \frac{3 a - 2 l + 3}{4 (l - 1)} (l \geq 2); \\ ((2 l - 1) n - \frac{3 a - 2 l + 7}{2}) S_{a, n + 1} + \frac{a - 2 l + 3}{2} n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 4}{4 (l - 1)} \leq n \leq \frac{3 a - 2 l + 7}{2 (2 l - 3)} (l \geq 2); \\ (2 n) S_{a, n + 1} + \frac{a - 1}{2} n S_{a, n + 2} - S_{a, n} \\ if n \geq \frac{3 a + 5}{2} . \end{matrix} \end{matrix}

Remark 2.

It is clear that there is no integer between

\frac{3 a}{4 l} - \frac{l - 3}{2 l}

and

\frac{3 a}{4 l} - \frac{l - 2}{2 l}

(exclusive), and between

\frac{3 a}{2 (2 l - 1)} - \frac{l - 1}{2 l - 1}

and

\frac{3 a}{2 (2 l - 1)}

(exclusive).

More precisely, for example, when

a = 8

and

a = 9

we have

\begin{matrix} g_{0} (S_{8, n}, S_{8, n + 1}, S_{8, n + 2}) \\ = \{\begin{matrix} 4 S_{8, n + 1} + 2 S_{8, n + 2} - S_{8, n} & if n = 2; \\ 6 S_{8, n + 1} + 6 S_{8, n + 2} - S_{8, n} & if n = 3; \\ (4 n - 13) S_{8, n + 1} + 3 n S_{8, n + 2} - S_{8, n} & if 4 \leq n \leq 5; \\ 2 n S_{8, n + 1} + 3 n S_{8, n + 2} - S_{8, n} & if 7 \leq n \leq 12; \\ (2 n - 14) S_{8, n + 1} + 4 n S_{8, n + 2} - S_{8, n} & if n \geq 13 \end{matrix} \end{matrix}

and

\begin{matrix} g_{0} (S_{9, n}, S_{9, n + 1}, S_{9, n + 2}) \\ = \{\begin{matrix} S_{9, 3} + 4 S_{9, 4} - S_{9, 2} & if n = 2; \\ 6 S_{9, 4} + 6 S_{9, 5} - S_{9, 3} & if n = 3; \\ 6 S_{9, 5} + 12 S_{9, 6} - S_{9, 4} & if n = 4; \\ 2 n S_{9, n + 1} + 3 n S_{9, n + 2} - S_{9, n} & if 5 \leq n \leq 6; \\ (3 n - 15) S_{9, n + 1} + 4 n S_{9, n + 2} - S_{9, n} & if 7 \leq n \leq 15; \\ 2 n S_{9, n + 1} + 4 n S_{9, n + 2} - S_{9, n} & if n \geq 16, \end{matrix} \end{matrix}

respectively.

4.1. Proof of Theorem 1

4.1.1. Even Case

Let a be even with

a \geq 4

. For a positive integer l, we shall prove that the elements of the 0-Apéry set can be arranged as in Table 1. Here, for simplicity, consider the representation

t_{y, z} : = y S_{a, n + 1} + z S_{a, n + 2} (y, z \geq 0) .

First, from the length relationship in the horizontal direction (y value) in Table 1,

2 l n - \frac{3}{2} a + l - 3 \geq 0 and 2 l n - \frac{3}{2} a + l - 3 \leq 2 n,

\frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)} .

Since

S_{a, 1} = 1

, we must assume that

n \geq 2

. Hence,

\frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)} \geq 2 or a \geq \frac{10 l - 14}{3} .

(4)

Since

t_{2 n + 1, 0} - t_{0, n} = (n + 1) S_{a, n}

t_{2 n + 1, 0} \equiv t_{0, n} (mod S_{a, n}) and t_{2 n + 1, 0} > t_{0, n} .

(5)

Since

t_{2 l n - \frac{3 a}{2} + l - 2, 0} + t_{0, (\frac{a}{2} - l) n + 1} = \frac{(a + 2 l) (n + 1) - 2}{2} S_{a, n},

we have

t_{2 l n - \frac{3 a}{2} + l - 2, 0} \equiv - t_{0, (\frac{a}{2} - l) n + 1} (mod S_{a, n}) .

(6)

Now, we need to see that the sequence

{ℓ S_{a, n + 1}}_{ℓ = 0}^{S_{a, n} - 1}

runs over every element of the 0-Apéry set once and all. After the long subsequence with length

2 n + 1

t_{0, j}, t_{1, j}, \dots, t_{2 n, j} (j = 0, 1, \dots, (\frac{a}{2} - l) n),

by (5), it is continued to the next subsequence by increasing by n rows:

t_{0, j + n}, t_{1, j + n}, \dots .

After the short subsequence with length

2 l n - \frac{3 a}{2} + l - 2

t_{0, j}, t_{1, j}, \dots, t_{2 l n - \frac{3 a}{2} + l - 3, j} (j = (\frac{a}{2} - l) n + 1, (\frac{a}{2} - l) n + 2, \dots, (\frac{a}{2} - l + 1) n),

by (6), it is continued to the long subsequence by decreasing by (

(\frac{a}{2} - l) n + 1

) rows:

t_{0, j - (\frac{a}{2} - l) n - 1}, t_{1, j - (\frac{a}{2} - l) n - 1}, \dots, t_{2 n, j - (\frac{a}{2} - l) n - 1} .

Since

gcd (n, (\frac{a}{2} - l) n + 1) = 1

, by this process, any element is not overlapped, and all the elements can be counted only once. Since

gcd (S_{a, n}, S_{a, n + 1}) = 1

, we have

{ℓ S_{a, n + 1}}_{ℓ = 0}^{S_{a, n} - 1} = {ℓ}_{ℓ = 0}^{S_{a, n} - 1}

In Table 1, there are two candidates to take the largest element of the 0-Apéry set:

t_{2 n, (\frac{a}{2} - l) n}

t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l + 1) n}

. We see that

\begin{matrix} t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l + 1) n} - t_{2 n, (\frac{a}{2} - l) n} : = E (a, n, l) \\ = (2 l - 1) a n^{3} - (\frac{3 a^{2}}{2} - (3 l - 2) a) n^{2} - (\frac{3 a^{2}}{2} - (l - 1) a - 2 l + 1) n \\ - \frac{3 a}{2} + l - 3 . \end{matrix}

Consider the largest root of

E (a, n, l) = 0

on n. Since for

l \geq 2

and

a \geq \frac{10 l - 14}{3}

in (4)

\begin{matrix} E (a, \frac{3 a - 2 l + 3}{2 (2 l - 1)}, l) & = \frac{a (3 a + 2 l + 1) (3 a - 2 l + 3)}{8 {(2 l - 1)}^{2}} - \frac{3}{2} \\ \geq \frac{480 l^{3} - 1924 l^{2} + 2439 l - 1019}{12 {(2 l - 1)}^{2}} > 0, \\ E (a, \frac{3 a - 2 l + 2}{2 (2 l - 1)}, l) & = - 2 < 0, \end{matrix}

when

n \geq \frac{3 a - 2 l + 3}{2 (2 l - 1)}

t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l + 1) n}

is the largest, and by Lemma 1 (2), we have

\begin{matrix} g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = (2 l n - \frac{3}{2} a + l - 3) S_{a, n + 1} + (\frac{a}{2} - l + 1) n S_{a, n + 2} - S_{a, n} . \end{matrix}

When

n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)}

t_{2 n, (\frac{a}{2} - l) n}

is the largest, and we have

g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n) S_{a, n + 1} + (\frac{a}{2} - l) n S_{a, n + 2} - S_{a, n} .

Since for

l = 1

and

a \geq 2

\begin{matrix} E (a, \frac{3 a}{2} + 1, 1) & = \frac{9 a^{3}}{4} + \frac{9 a^{2}}{2} + 2 a - 1 > 0, \\ E (a, \frac{3 a}{2}, 1) & = - 2 < 0, \end{matrix}

when

n \geq \frac{3 a}{2} + 1

t_{2 n - \frac{3}{2} a - 2, \frac{a}{2} n}

is the largest, and we have

g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n - \frac{3}{2} a - 2) S_{a, n + 1} + \frac{a}{2} n S_{a, n + 2} - S_{a, n} .

When

n \leq \frac{3 a}{2}

t_{2 n, (\frac{a}{2} - 1) n}

is the largest, and we have

g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n) S_{a, n + 1} + (\frac{a}{2} - 1) n S_{a, n + 2} - S_{a, n} .

4.1.2. Odd Case

Let a be odd with

a \geq 5

. We shall prove that when

n \leq 3 (a + 1) / 2

, the elements of the 0-Apéry set can be arranged as in Table 2. First, from the length relationship in the horizontal direction (y value) in Table 2,

(2 l - 1) n - \frac{3 a - 2 l + 7}{2} \geq 0 and (2 l - 1) n - \frac{3 a - 2 l + 7}{2} \leq 2 n,

\frac{3 a - 2 l + 7}{2 (2 l - 1)} \leq n \leq \frac{3 a - 2 l + 7}{2 (2 l - 3)} .

Since the possible

n \geq 2

, we have

n \geq \frac{3 a - 2 l + 7}{2 (2 l - 3)} or a \geq \frac{10 l - 19}{3} .

(7)

Since

t_{(2 l - 1) n - \frac{3 a - 2 l + 5}{2}, 0} + t_{0, \frac{a - 2 l + 1}{2} n + 1} = \frac{(a + 2 l) (n + 1) - n - 3}{2} S_{a, n},

we have

t_{(2 l - 1) n - \frac{3 a - 2 l + 5}{2}, 0} \equiv - t_{0, \frac{a - 2 l + 1}{2} n + 1} (mod S_{a, n}) .

(8)

Now, the sequence

{ℓ S_{a, n + 1}}_{ℓ = 0}^{S_{a, n} - 1}

runs over the elements of 0-Apéry set once and all. After the long subsequence with length

2 n + 1

t_{0, j}, t_{1, j}, \dots, t_{2 n, j} (j = 0, 1, \dots, \frac{a - 2 l + 1}{2} n),

by (5), it is continued to the next subsequence by increasing by n rows:

t_{0, j + n}, t_{1, j + n}, \dots .

After the short subsequence with length

(2 l - 1) n - \frac{3 a - 2 l + 5}{2}

t_{0, j}, t_{1, j}, \dots, t_{(2 l - 1) n - \frac{3 a - 2 l + 5}{2}, j} (j = \frac{a - 2 l + 1}{2} n + 1, \frac{a - 2 l + 1}{2} n + 2, \dots, \frac{a - 2 l + 3}{2} n),

by (8), it is continued to the long subsequence by decreasing by (

\frac{a - 2 l + 1}{2} n + 1

) rows:

t_{0, j - \frac{a - 2 l + 1}{2} n - 1}, t_{1, j - \frac{a - 2 l + 1}{2} n - 1}, \dots, t_{2 n, j - \frac{a - 2 l + 1}{2} n - 1} .

Since

gcd (n, \frac{a - 2 l + 1}{2} n + 1) = 1

, by this process, any element is not overlapped, and all the elements can be counted only once. Since

gcd (S_{a, n}, S_{a, n + 1}) = 1

, we have

{ℓ S_{a, n + 1}}_{ℓ = 0}^{S_{a, n} - 1} = {ℓ}_{ℓ = 0}^{S_{a, n} - 1}

In Table 2, there are two candidates to take the largest values:

t_{2 n, \frac{a - 2 l + 1}{2} n}

and

t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 3}{2} n}

. Notice that

\begin{matrix} t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 3}{2} n} - t_{2 n, \frac{a - 2 l + 1}{2} n} : = O (a, n, l) \\ = 2 (l - 1) a n^{3} - \frac{a (3 a - 6 l + 7)}{2} n^{2} - \frac{3 a^{2} - (2 l - 3) a - 4 (l - 1)}{2} n \\ - \frac{3 a - 2 l + 7}{2} . \end{matrix}

Since for

l \geq 2

and

a \geq \frac{10 l - 19}{3}

in (7)

\begin{matrix} O (a, \frac{3 a - 2 l + 4}{4 (l - 1)}, l) & = \frac{a (3 a + 2 l) (3 a - 2 l + 4)}{32 {(l - 1)}^{2}} - \frac{3}{2} \\ \geq \frac{960 l^{3} - 5288 l^{2} + 9467 l - 5559}{96 {(l - 1)}^{2}} > 0, \\ O (a, \frac{3 a - 2 l + 3}{4 (l - 1)}, l) & = - 2 < 0, \end{matrix}

when

n \leq \frac{3 a - 2 l + 3}{4 (l - 1)}

t_{2 n, \frac{a - 2 l + 1}{2} n}

is the largest in the 0-Apéry set. Hence, by Lemma 1 (2), we have

g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n) S_{a, n + 1} + \frac{a - 2 l + 1}{2} n S_{a, n + 2} - S_{a, n} .

When

n \geq \frac{3 a - 2 l + 4}{4 (l - 1)}

t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 3}{2} n}

is the largest, and we have

\begin{matrix} g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = ((2 l - 1) n - \frac{3 a - 2 l + 7}{2}) S_{a, n + 1} + \frac{a - 2 l + 3}{2} n S_{a, n + 2} - S_{a, n} . \end{matrix}

For

l = 1

, by

O (a, n, 1) = - \frac{n (n + 1) a (3 a + 1)}{2} - \frac{3 a + 5}{2} < 0,

we can find that

t_{2 n, \frac{a - 1}{2} n}

is the largest in the 0-Apéry set. Hence, by Lemma 1 (2), we have

g_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n) S_{a, n + 1} + \frac{a - 1}{2} n S_{a, n + 2} - S_{a, n} .

5. $p = 1$

The arrangement of the elements of the 1-Apéry set can be determined from those of the 0-Apéry set. As seen in the case where

p = 0

, there are four different patterns about the arrangement of the elements of the 0-Apéry set.

Consider the case where a is even with

a \geq 4

. Otherwise, the following argument is not valid. By (5), (6) and

t_{- \frac{3 a}{2} + 2 n - 1, (\frac{a}{2} - 1) n + 1} = \frac{(a + 2) n + a}{2} S_{a, n}

, we have the correspondences between the elements of the 0-Apéry set and those of the 1-Apéry set:

\begin{matrix} t_{y, z} \equiv t_{y + 2 n + 1, z - n} (mod S_{a, n}) (0 \leq y \leq 2 n, n \leq z \leq (\frac{a}{2} - l) n; \\ 0 \leq y \leq 2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l) n + 1 \leq z \leq (\frac{a}{2} - l + 1) n), \\ t_{y, z} \equiv t_{y + 2 l n - \frac{3 a}{2} + l - 2, z + (\frac{a}{2} - l) n + 1} (mod S_{a, n}) \\ (0 \leq y \leq \frac{3 a}{2} - 2 (l - 1) n - l + 2, 0 \leq z \leq n - 1), \\ t_{y, z} \equiv t_{y - \frac{3 a}{2} + 2 (l - 1) n + l - 3, z + (\frac{a}{2} - l + 1) n + 1} (mod S_{n}) \\ (\frac{3 a}{2} - 2 (l - 1) n - l + 3 \leq y \leq 2 n, 0 \leq z \leq n - 1), \end{matrix}

respectively. Namely, in Table 3, the elements in the first n rows are simply moved below the 0-Apéry set to fill in the gap. However, the remaining portion is moved to the lower left. Elements other than the first n rows are shifted to the right side of the 0-Apéry set by shifting up n rows.

Set

τ_{x, y, x} : = x S_{a, n} + y S_{a, n + 1} + z S_{a, n + 2}

. We can show that all the elements of the 1-Apéry set have at least 2 different representations:

\begin{matrix} τ_{n + 1, y, z} = τ_{0, y + 2 n + 1, z - n} (0 \leq y \leq 2 n, n \leq z \leq (\frac{a}{2} - l) n; \\ 0 \leq y \leq 2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l) n + 1 \leq z \leq (\frac{a}{2} - l + 1) n), \\ τ_{\frac{(a + 2 l) (n + 1) - 2}{2}, y, z} = τ_{0, y + 2 l n - \frac{3 a}{2} + l - 2, z + (\frac{a}{2} - l) n + 1} \\ (0 \leq y \leq \frac{3 a}{2} - 2 (l - 1) n - l + 2, 0 \leq z \leq n - 1), \\ τ_{\frac{(a + 2 l - 2) n + a + 2 l - 4}{2}, y, z} = τ_{0, y - \frac{3 a}{2} + 2 (l - 1) n + l - 3, z + (\frac{a}{2} - l + 1) n + 1} \\ (\frac{3 a}{2} - 2 (l - 1) n - l + 3 \leq y \leq 2 n, 0 \leq z \leq n - 1) . \end{matrix}

From Table 3, there are four candidates to take the largest value of

{Ap}_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 2) n} t_{2 n, (\frac{a}{2} - l + 1) n}, t_{2 (l + 1) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l) n}, t_{4 n + 1, (\frac{a}{2} - l - 1) n} .

Since

\begin{matrix} t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 2) n} - t_{2 (l + 1) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l) n} \\ = t_{2 n, (\frac{a}{2} - l + 1) n} - t_{4 n + 1, (\frac{a}{2} - l - 1) n} = 3 a n (n + 1) - 1 > 0, \end{matrix}

similarly to the case where

p = 0

, there are only two possibilities. Namely, when

\frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)}

(

l \geq 2

) or

\frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}

(

l = 1

t_{2 n, (\frac{a}{2} - l + 1) n}

is the largest, so by Lemma 1 (2) we have

g_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = (2 n) S_{a, n + 1} + (\frac{a}{2} - l + 1) n S_{a, n + 2} - S_{a, n} .

When

\frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)}

(

l \geq 2

) or

n \geq \frac{3 a}{2} + 1

(

l = 1

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 2) n}

is the largest, and we have

\begin{matrix} g_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = (2 l n - \frac{3 a}{2} + l - 3) S_{a, n + 1} + (\frac{a}{2} - l + 2) n S_{a, n + 2} - S_{a, n} . \end{matrix}

The case where a is odd is not so similar, but analogous patterns and corresponding congruences and inequalities give the following results. When a is odd with

a \geq 5

and

n \geq 2

, we have

\begin{matrix} g_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + \frac{a - 2 l + 3}{2} n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 7}{2 (2 l - 1)} \leq n \leq \frac{3 a - 2 l + 3}{4 (l - 1)} (l \geq 2); \\ ((2 l - 1) n - \frac{3 a - 2 l + 7}{2}) S_{a, n + 1} + \frac{a - 2 l + 5}{2} n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 + 4}{4 (l - 1)} \leq n \leq \frac{3 a - 2 l + 7}{2 (2 l - 3)} (l \geq 2); \\ (2 n) S_{a, n + 1} + \frac{a + 1}{2} n S_{a, n + 2} - S_{a, n} \\ if n \geq \frac{3 a + 5}{2} . \end{matrix} \end{matrix}

6. $p \geq 2$

The elements of the 2-Apéry set can be determined from those of the 1-Apéry set as follows. See Table 4.

Similarly to the case where

p = 1

, the elements in the first n rows of the main part of the 1-Apéry set are simply moved below the 0-Apéry set to fill in the gap. However, the remaining portion is moved to the lower left of the 0-Apéry. Elements other than the first n rows of the main part are shifted to the right side of the 0-Apéry set by shifting up n rows. The other staircase parts are shifted to the left by

(2 n + 1)

and upward by n.

In Table 4, by comparing the six candidates

\begin{matrix} t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 3) n}, t_{2 n, (\frac{a}{2} - l + 2) n}, t_{2 (l + 1) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l + 1) n}, \\ t_{4 n + 1, (\frac{a}{2} - l) n}, t_{2 (l + 2) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l - 1) n}, t_{6 n + 2, (\frac{a}{2} - l - 2) n}, \end{matrix}

we find that

t_{2 n, (\frac{a}{2} - l + 2) n}

is the largest element of the 2-Apéry set when

\frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)}

(

l \geq 2

) or

\frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}

(

l = 1

);

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 3) n}

is the largest element of the 2-Apéry set when

\frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)}

(

l \geq 2

) or

n \geq \frac{3 a}{2} + 1

(

l = 1

). Therefore, by Lemma 1 (2) we have

\begin{matrix} g_{2} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + (\frac{a}{2} - l + 2) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)} (l \geq 2); \\ (2 l n - \frac{3 a}{2} + l - 3) S_{a, n + 1} + (\frac{a}{2} - l + 3) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)} (l \geq 2), \\ (2 n) S_{a, n + 1} + (\frac{a}{2} + 1) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}; \\ (2 n - \frac{3 a}{2} - 2) S_{a, n + 1} + (\frac{a}{2} + 2) n S_{a, n + 2} - S_{a, n} \\ if n \geq \frac{3 a}{2} + 1 . \end{matrix} \end{matrix}

Similarly, for

p = 3, 4, \dots

, the elements of the p-Apéry set can also be determined from those of the (

p - 1

)-Apéry set. However, the same pattern cannot be continued. As seen in Table 5, the situation is changed.

0_{a}

1_{a}

and

2_{a}

indicate the position of the largest elements of the 0-, 1- and 2-Apéry sets, respectively, when

\frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)}

(

l \geq 2

) or

n \geq \frac{3 a}{2} + 1

(

l = 1

); and

0_{b}

1_{b}

and

2_{b}

when

\frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)}

(

l \geq 2

) or

\frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}

(

l = 1

). We can see that no element of the Apéry set can exist at the expected location for the largest value. Then, in general, for

p \geq \frac{a}{2} - l + 1

, the same formula cannot be applied, and the situation becomes more and more complicated, though the p-Frobenius numbers should exist.

Therefore, for a general integer p with

0 \leq p \leq \frac{a}{2} - l

we have the following. Note that the case where a is odd is not so similar, but analogous arguments can be applied.

Theorem 2.

When a is even with

a \geq 6

and

n \geq 2

, for integers

0 \leq p \leq \frac{a}{2} - l

and

l \geq 2

\begin{matrix} g_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + (p + \frac{a}{2} - l) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a}{4 l} - \frac{l - 3}{2 l} \leq n \leq \frac{3 a - 2 l + 2}{2 (2 l - 1)}; \\ (2 l n - \frac{3 a}{2} - 1) S_{a, n + 1} + (p + \frac{a}{2} - l + 1) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 3}{2 (2 l - 1)} \leq n \leq \frac{3 a}{4 (l - 1)} - \frac{l - 3}{2 (l - 1)}, \end{matrix} \end{matrix}

for integers

n \geq 2

and

0 \leq p \leq \frac{a}{2} - 1

\begin{matrix} g_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + (p + \frac{a}{2} - 1) n S_{a, n + 2} - S_{a, n} & if \frac{3 a}{4} + 1 \leq n \leq \frac{3 a}{2}; \\ (2 n - \frac{3 a}{2} - 2) S_{a, n + 1} + (p + \frac{a}{2}) n S_{a, n + 2} - S_{a, n} & if n \geq \frac{3 a}{2} + 1 . \end{matrix} \end{matrix}

When a is odd with

a \geq 5

and

n \geq 2

, for

0 \leq p \leq \frac{a - 2 l + 1}{2}

with

l \geq 2

\begin{matrix} g_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \{\begin{matrix} (2 n) S_{a, n + 1} + (p + \frac{a - 2 l + 1}{2}) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 7}{2 (2 l - 1)} \leq n \leq \frac{3 a - 2 l + 3}{4 (l - 1)}; \\ ((2 l - 1) n - \frac{3 a - 2 l + 7}{2}) S_{a, n + 1} + (p + \frac{a - 2 l + 3}{2}) n S_{a, n + 2} - S_{a, n} \\ if \frac{3 a - 2 l + 4}{4 (l - 1)} \leq n \leq \frac{3 a - 2 l + 7}{2 (2 l - 3)}, \end{matrix} \end{matrix}

for integers

n \geq 2

and

0 \leq p \leq \frac{a - 1}{2}

\begin{matrix} g_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = (2 n) S_{a, n + 1} + (p + \frac{a - 1}{2}) n S_{a, n + 2} - S_{a, n} if n \geq \frac{3 a + 5}{2} . \end{matrix}

Remark 3.

When a is even, the first and second conditions in the first identity are equivalent to

\frac{3 a + 6}{4 n + 2} \leq l \leq \frac{3 a + 2 n + 2}{4 n + 2} and \frac{3 a + 2 n + 3}{4 n + 2} \leq l \leq \frac{3 a + 4 n + 6}{4 n + 2},

respectively. When a is odd, the first and second conditions in the third identity are equivalent to

\frac{3 a + 2 n + 7}{4 n + 2} \leq l \leq \frac{3 a + 4 n + 3}{4 n + 2} and \frac{3 a + 4 n + 4}{4 n + 2} \leq l \leq \frac{3 a + 6 n + 7}{4 n + 2},

respectively.

7. p-Genus

Let a be even with

a \geq 4

. Observing Table 1, Table 3 and Table 4, the sum of the elements of the p-Apéry set is made by dividing into three parts: all the left sides of the staircase, all the right sides of the staircase, and the main part. For

0 \leq p \leq \frac{a}{2} - l

with

l \geq 1

we have

\begin{matrix} \sum_{j = 0}^{S_{a, n} - 1} m_{j}^{(p)} \\ = \sum_{κ = 0}^{p - 1} \sum_{y = (2 n + 1) κ}^{(2 n + 1) κ + 2 l n - \frac{3 a}{2} + l - 3} \sum_{z = (p + \frac{a}{2} - 2 κ - l) n + 1}^{(p + \frac{a}{2} - 2 κ - l + 1) n} t_{y, z} \\ + \sum_{κ = 0}^{p - 1} \sum_{y = (2 n + 1) κ + 2 l n - \frac{3 a}{2} + l - 2}^{(2 n + 1) (κ + 1) - 1} \sum_{z = (p + \frac{a}{2} - 2 κ - l - 1) n + 1}^{(p + \frac{a}{2} - 2 κ - l) n} t_{y, z} \\ + \sum_{y = p (2 n + 1)}^{p (2 n + 1) + 2 n} \sum_{z = 0}^{(\frac{a}{2} - p - l) n} t_{y, z} + \sum_{y = p (2 n + 1)}^{p (2 n + 1) + 2 l n - \frac{3 a}{2} + l - 3} \sum_{z = (\frac{a}{2} - p - l) n + 1}^{(\frac{a}{2} - p - l + 1) n} t_{y, z} \\ = \frac{S_{a, n}}{8} (n (2 (a^{2} + 4 a + 4 l (l - 1)) n^{2} + (3 a^{2} - 2 (6 l - 5) a + 4 (3 l - 1) (l - 2)) n \\ + a^{2} - 2 (6 l - 5) a + 4 (l - 2) (l - 3)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

By Lemma 1 (3), we have

\begin{matrix} n_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) = \frac{1}{S_{a, n}} \sum_{j = 0}^{S_{a, n} - 1} m_{j}^{(p)} - \frac{S_{a, n} - 1}{2} \\ = \frac{1}{8} (n (2 (a^{2} + 4 a + 4 l (l - 1)) n^{2} + (3 a^{2} - 6 (2 l - 1) a + 4 (3 l - 1) (l - 2)) n \\ + a^{2} - 2 (6 l - 7) a + 4 (l - 2) (l - 3)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

When a is odd with

a \geq 5

, for

0 \leq p \leq \frac{a - 2 l + 1}{2}

with

l \geq 1

we have

\begin{matrix} \sum_{j = 0}^{S_{a, n} - 1} m_{j}^{(p)} \\ = \sum_{κ = 0}^{p - 1} \sum_{y = (2 n + 1) κ}^{(2 n + 1) κ + (2 l - 1) n - \frac{3 a - 2 l + 7}{2}} \sum_{z = (p + \frac{a - 2 l + 1}{2} - 2 κ) n + 1}^{(p + \frac{a - 2 l + 3}{2} - 2 κ) n} t_{y, z} \\ + \sum_{κ = 0}^{p - 1} \sum_{y = (2 n + 1) κ + (2 l - 1) n - \frac{3 a - 2 l + 5}{2}}^{(2 n + 1) (κ + 1) - 1} \sum_{z = (p + \frac{a - 2 l - 1}{2} - 2 κ) n + 1}^{(p + \frac{a - 2 l + 1}{2} - 2 κ) n} t_{y, z} \\ + \sum_{y = p (2 n + 1)}^{p (2 n + 1) + 2 n} \sum_{z = 0}^{(\frac{a - 2 l + 1}{2} - p) n} t_{y, z} + \sum_{y = p (2 n + 1)}^{p (2 n + 1) + (2 l - 1) n - \frac{3 a - 2 l + 7}{2}} \sum_{z = (\frac{a - 2 l + 1}{2} - p) n + 1}^{(\frac{a - 2 l + 3}{2} - p) n} t_{y, z} \\ = \frac{S_{a, n}}{8} (n (2 ((a + 1) (a + 3) + 4 l (l - 2)) n^{2} + (3 a^{2} - 4 (3 l - 4) a + (2 l - 5) (6 l - 5)) n \\ + a^{2} - 4 (3 l - 4) a + (2 l - 5) (2 l - 7)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

By Lemma 1 (3), we have

\begin{matrix} n_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \frac{1}{8} (n (2 ((a + 1) (a + 3) + 4 l (l - 2)) n^{2} + (3 a^{2} - 12 (l - 1) a + (2 l - 5) (6 l - 5)) n \\ + a^{2} - 4 (3 l - 5) a + (2 l - 5) (2 l - 7)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

Theorem 3.

When a is even with

a \geq 4

and

n \geq 2

, for

0 \leq p \leq \frac{a}{2} - l

with

l \geq 1

\begin{matrix} n_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \frac{1}{8} (n (2 (a^{2} + 4 a + 4 l (l - 1)) n^{2} + (3 a^{2} - 6 (2 l - 1) a + 4 (3 l - 1) (l - 2)) n \\ + a^{2} - 2 (6 l - 7) a + 4 (l - 2) (l - 3)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

When a is odd with

a \geq 5

and

n \geq 2

, for

0 \leq p \leq \frac{a - 2 l + 1}{2}

with

l \geq 1

\begin{matrix} n_{p} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2}) \\ = \frac{1}{8} (n (2 ((a + 1) (a + 3) + 4 l (l - 2)) n^{2} + (3 a^{2} - 12 (l - 1) a + (2 l - 5) (6 l - 5)) n \\ + a^{2} - 4 (3 l - 5) a + (2 l - 5) (2 l - 7)) \\ + 4 (2 n + 1) (2 S_{a, n + 1} - n (n + 1)) p - 4 n (n + 1) (2 n + 1) p^{2}) . \end{matrix}

8. Examples

When

(a, n) = (22, 9)

, by Remark of Theorem 2, we see that

2 \in [\frac{36}{19}, \frac{43}{19}]

and there is no integer in the interval

[\frac{87}{38}, \frac{54}{19}]

, the first identity of the even case in Theorem 2 is applied with

l = 2

. Then, for

0 \leq p \leq 9 = \frac{22}{2} - 2

\begin{matrix} g_{p} (S_{22, 9}, S_{22, 10}, S_{22, 11}) = g_{p} (1585, 1981, 2421) \\ = 18 S_{22, 10} + 22 (p + 9) S_{22, 11} - S_{22, 10} . \end{matrix}

The first several terms are

\begin{matrix} {g_{p} (S_{22, 9}, S_{22, 10}, S_{22, 11})}_{p = 0}^{9} = 230174, 251963, 273752, 295541, \\ 317330, 339119, 360908, 382697, 404486, 426275 . \end{matrix}

However, when

p = 10

, the identity does not give the correct value of 442125 but the wrong value of 448064.

Concerning the p-genus, by the first identity of Theorem 3, we have

\begin{matrix} {n_{p} (S_{22, 9}, S_{22, 10}, S_{22, 11})}_{p = 0}^{9} = 116694, 152623, 186842, 219351, \\ 250150, 279239, 306618, 332287, 356246, 378495 . \end{matrix}

When

(a, n) = (13, 11)

, by Remark of Theorem 2, we see that there is no integer in the interval

[\frac{34}{23}, \frac{43}{23}]

and

2 \in [\frac{87}{46}, \frac{56}{23}],

the second identity of the odd case in Theorem 2 is applied with

l = 2

. Then, for

0 \leq p \leq 5 = \frac{13 - 3}{2}

we have

\begin{matrix} {g_{p} (S_{13, 11}, S_{13, 12}, S_{13, 13})}_{p = 0}^{5} \\ = 153087, 175406, 197725, 220044, 242363, 264682 . \end{matrix}

Concerning the p-genus, by the third identity of Theorem 3, we have

\begin{matrix} {n_{p} (S_{13, 11}, S_{13, 12}, S_{13, 13})}_{p = 0}^{5} \\ = 79904, 116359, 149778, 180161, 207508, 231819 . \end{matrix}

9. Final Comments

One should not think that the results in this paper are quite similar to some previous works. In general, it is never easy to find any explicit formula for the Frobenius number for given sequences or tuples with three or more variables. When

p > 0

, the situation is even harder. For example, triangular numbers are given as a polynomial of similar quadratic polynomials too. The explicit formulas of the p-Frobenius number of consecutive triangular numbers have been successful in being given in [14,19] and the case of squares is also possible [26]. But nothing has been known for pentagonal, hexagonal, heptagonal, octagonal numbers given by

n (k n - k + 2) / 2

for

k = 5, 6, 7, 8

, respectively.

Author Contributions

Writing—original draft preparation, T.K.; writing—review and editing, R.Y., J.M.; All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflict of interest.

References

Matone M.; Volpato, R. Vector-valued modular forms from the Mumford forms, Schottky-Igusa form, product of Thetanullwerte and the amazing Klein formula. Proc. Amer. Math. Soc. 2013, 141, 2575–2587.
Plouffe, S. Approximations of generating functions and a few conjectures. Dissertation, Universit’e du Québec à Montréal, 1992; arXiv:0911.4975v1 (2009).
N. J. A. Sloane, The on-line encyclopedia of integer sequences, available at oeis.org. (2024).
Komatsu, T.; Ying, H. p-numerical semigroups with p-symmetric properties. J. Algebra Appl. 2024, (Online ready), 2450216. [CrossRef]
D. S. Binner, Generalization of a result of Sylvester related to the Frobenius coin problem, J. Integer Seq. 2021, 24, no. 8, Art. 21.8.4, 14 pp.
Cayley, A. On a problem of double partitions. Philos. Mag. 1860, 20, 337–341.
Komatsu, T. On the number of solutions of the Diophantine equation of Frobenius–General case. Math. Commun. 2003, 8, 195–206.
Tripathi, A. The number of solutions to ax + by = n. Fibonacci Quart. 2000, 38, 290–293.
A. Assi, M. D’Anna and P. A. Garcia-Sanchez, Numerical semigroups and applications, 2nd extended and revised edition, RSME Springer Series 3. Cham: Springer (2020).
Rosales, J.C.; Garcia-Sanchez, P.A. Numerical semigroups. Developments in Mathematics, 20. Springer, New York, 2009.
Curtis, F. On formulas for the Frobenius number of a numerical semigroup. Math. Scand. 1990, 67, 190–192.
Rosales, J.C.; Branco, M.B.; Torrão, D. The Frobenius problem for repunit numerical semigroups. Ramanujan J. 2016, 40, 323–334. [CrossRef]
Rosales, J.C.; Branco, M.B.; Torrão, D. The Frobenius problem for Mersenne numerical semigroups. Math. Z. 2017, 286, 741–749. [CrossRef]
Robles-Perez, A.M.; Rosales, J.C. The Frobenius number for sequences of triangular and tetrahedral numbers. J. Number Theory 2018, 186, 473–492. [CrossRef]
Komatsu, T. The Frobenius number associated with the number of representations for sequences of repunits. C. R. Math. Acad. Sci. Paris 2023, 361, 73–89. [CrossRef]
Komatsu, T.; Ying, H. The p-Frobenius and p-Sylvester numbers for Fibonacci and Lucas triplets. Math. Biosci. Eng. 2023, 20, 3455–3481. [CrossRef]
Komatsu, T.; Mu, J. p-numerical semigroups of Pell triples. J. Ramanujan Math. Soc. (in press).https://jrms.ramanujanmathsociety.org/articles_in_press.html.
Komatsu, T.; Pita-Ruiz, C. The Frobenius number for Jacobsthal triples associated with number of solutions. Axioms 2023, 12, 98. [CrossRef]
Komatsu, T. The Frobenius number for sequences of triangular numbers associated with number of solutions. Ann. Comb. 2022, 26, 757–779. [CrossRef]
Aigner, M.; Ziegler, G.M. Proofs from THE BOOK. Fifth edition (Including illustrations by Karl H. Hofmann), Springer-Verlag, Berlin, 2014, ISBN: 978-3-662-44204-3; 978-3-662-44205-0.
Apéry, R. Sur les branches superlinéaires des courbes algébriques. C. R. Acad. Sci. Paris 1946, 222, 1198–1200.
Komatsu, T. On the determination of p-Frobenius and related numbers using the p-Apéry set. Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 2024, 118, 58. [CrossRef]
Brauer, A.; Shockley, B.M. On a problem of Frobenius. J. Reine. Angew. Math. 1962, 211, 215–220. [CrossRef]
Selmer, E.S. On the linear diophantine problem of Frobenius. J. Reine Angew. Math. 1977, 293–294, 1–17. [CrossRef]
Komatsu, T. Sylvester power and weighted sums on the Frobenius set in arithmetic progression. Discret. Appl. Math. 2022, 315, 110–126. [CrossRef]
Lepilov, M.; O’Rourke, J.; Swanson, I. Frobenius numbers of numerical semigroups generated by three consecutive squares or cubes. Semigroup Forum 2015, 91, No.1, 238–259. [CrossRef]

Figure 1. Chinese Checker.

Table 1.

{Ap}_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is even.

Table 1.

{Ap}_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is even.

$t_{0, 0}$	⋯	$t_{2 l n - \frac{3}{2} a + l - 3, 0}$	$t_{2 l n - \frac{3}{2} a + l - 2, 0}$	⋯	⋯	$t_{2 n, 0}$
⋮		⋮	⋮			⋮
$t_{0, (\frac{a}{2} - l) n}$	⋯	$t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l) n}$	$t_{2 l n - \frac{3}{2} a + l - 2, (\frac{a}{2} - l) n}$	⋯	⋯	$t_{2 n, (\frac{a}{2} - l) n}$
$t_{0, (\frac{a}{2} - l) n + 1}$	⋯	$t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l) n + 1}$
⋮		⋮
⋮		⋮
$t_{0, (\frac{a}{2} - l + 1) n}$	⋯	$t_{2 l n - \frac{3}{2} a + l - 3, (\frac{a}{2} - l + 1) n}$

Table 2.

{Ap}_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is odd.

Table 2.

{Ap}_{0} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is odd.

$t_{0, 0}$	⋯	$t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}), 0}$	$t_{(2 l - 1) n - \frac{3 a - 2 l + 5}{2}, 0}$	⋯	⋯	$t_{2 n, 0}$
⋮		⋮	⋮			⋮
$t_{0, \frac{a - 2 l + 1}{2} n}$	⋯	$t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 1}{2} n}$	$t_{(2 l - 1) n - \frac{3 a - 2 l + 5}{2}, \frac{a - 2 l + 1}{2} n}$	⋯	⋯	$t_{2 n, \frac{a - 2 l + 1}{2} n}$
$t_{0, \frac{a - 2 l + 1}{2} n + 1}$	⋯	$t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 1}{2} n + 1}$
⋮		⋮
⋮		⋮
$t_{0, \frac{a - 2 l + 3}{2} n}$	⋯	$t_{(2 l - 1) n - \frac{3 a - 2 l + 7}{2}, \frac{a - 2 l + 3}{2} n}$

Table 3.

{Ap}_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is even.

Table 3.

{Ap}_{1} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

when a is even.

t_{2 n + 1, 0}

⋯

t_{4 n + 1, 0}

⋮

t_{2 n + 1, (\frac{a}{2} - l - 1) n}

⋯

t_{4 n + 1, (\frac{a}{2} - l - 1) n}

t_{2 n + 1, (\frac{a}{2} - l - 1) n + 1}

⋯

t_{2 (l + 1) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l - 1) n + 1}

⋮

t_{2 n + 1, (\frac{a}{2} - l) n}

⋯

t_{2 (l + 1) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l) n}

t_{2 l n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l) n + 1}

⋯

t_{2 n, (\frac{a}{2} - l) n + 1}

⋮

t_{2 l n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l + 1) n}

⋯

t_{2 n, (\frac{a}{2} - 1) n}

t_{0, (\frac{a}{2} - 1) n + 1}

⋯

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 1) n + 1}

⋮

t_{0, (\frac{a}{2} - l + 2) n}

⋯

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 2) n}

Table 4.

{Ap}_{2} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

Table 4.

{Ap}_{2} (S_{a, n}, S_{a, n + 1}, S_{a, n + 2})

t_{4 n + 2, 0}

⋯

t_{(2 l + 4) n - \frac{3 a}{2} + l - 1, 0}

t_{(2 l + 4) n - \frac{3 a}{2} + l, 0}

⋯

t_{6 n + 2, 0}

⋮

t_{4 n + 2, (\frac{a}{2} - l - 2) n}

⋯

t_{(2 l + 4) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l - 2) n}

t_{(2 l + 4) n - \frac{3 a}{2} + l, (\frac{a}{2} - l - 2) n}

⋯

t_{6 n + 2, (\frac{a}{2} - l - 2) n}

t_{4 n + 2, (\frac{a}{2} - l - 2) n + 1}

⋯

t_{(2 l + 4) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l - 2) n + 1}

⋮

t_{4 n + 2, (\frac{a}{2} - l - 1) n}

⋯

t_{(2 l + 4) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l - 1) n}

t_{(2 l + 2) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l - 1) n + 1}

⋯

t_{4 n + 1, (\frac{a}{2} - l - 1) n + 1}

⋮

t_{(2 l + 2) n - \frac{3 a}{2} + l - 1, (\frac{a}{2} - l) n}

⋯

t_{4 n + 1, (\frac{a}{2} - l) n}

t_{2 n + 1, (\frac{a}{2} - l) n + 1}

⋯

t_{(2 l + 2) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l) n + 1}

⋮

t_{2 n + 1, (\frac{a}{2} - l + 1) n}

⋯

t_{(2 l + 2) n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l + 1) n}

t_{2 l n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l + 1) n + 1}

⋯

t_{2 n, (\frac{a}{2} - l + 1) n + 1}

⋮

t_{2 l n - \frac{3 a}{2} + l - 2, (\frac{a}{2} - l + 2) n}

⋯

t_{2 n, (\frac{a}{2} - l + 2) n}

t_{0, (\frac{a}{2} - l + 2) n + 1}

⋯

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 2) n + 1}

⋮

t_{0, (\frac{a}{2} - l + 3) n}

⋯

t_{2 l n - \frac{3 a}{2} + l - 3, (\frac{a}{2} - l + 3) n}

Table 5.

{Ap}_{3} (S_{n, a}, S_{n + 1, a}, S_{n + 2, a})

Table 5.

{Ap}_{3} (S_{n, a}, S_{n + 1, a}, S_{n + 2, a})

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

MDPI Initiatives

Important Links

Choose an area of interest and we will send you notifications of new preprints at your preferred frequency.

Disclaimer

The p-Frobenius Number for the Triple of the Generalized Star Numbers

Abstract

1. Introduction

2. Basic Properties of the Generalized Star Numbers

3. Apéry Set

4. Main Result

4.1. Proof of Theorem 1

4.1.1. Even Case

4.1.2. Odd Case

5. p = 1

6. p ≥ 2

7. p-Genus

8. Examples

9. Final Comments

Author Contributions

Funding

Data Availability Statement

Conflicts of Interest

References

MDPI Initiatives

Important Links

Subscribe

5. $p = 1$

6. $p \geq 2$