The Frobenius Number for Jacobsthal Triples Associated with Number of Solutions

Abstract

In this paper, we find a formula for the largest integer (p-Frobenius number) such that a linear equation of non-negative integer coefficients composed of a Jacobsthal triplet has at most p representations. For $p=0$, the problem is reduced to the famous linear Diophantine problem of Frobenius, the largest integer of which is called the Frobenius number. We also give a closed formula for the number of non-negative integers (p-genus), such that linear equations have at most p representations. Extensions to the Jacobsthal polynomial and the Jacobsthal–Lucas polynomial give more general formulas that include the familiar Fibonacci and Lucas numbers. A basic problem with the Fibonacci triplet was dealt by Marin, Ramírez Alfonsín and M. P. Revuelta for $p=0$ and by Komatsu and Ying for the general non-negative integer p.

1. Introduction

The Jacobsthal sequence ${\left\{J_n\right\}}_{n\ge 0}$ is defined by

$$J_n=J_{n-1}+2J_{n-2}(n\ge 2)\mathrm{with}{J}_0=0\mathrm{and}{J}_1=1$$

and the first terms are given by 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10,923, 21,845, 43,691, …([1] and ([2], A001045)). Jacobsthal [3] may never have seen the actual values of this sequence. However, Horadam [4] uses the name “Jacobsthal sequence”; such an important sequence needs a name. Many properties and identities can be found in ([1], Ch.44).

In this paper, for fixed integers $i,k\ge 3$, we consider the linear Diophantine equation of the form

$$J_i{x}_1+J_{i+2}{x}_2+J_{i+k}{x}_3=N.$$

(1)

For example, when $(i,k)=(4,5)$ and $N=268$, the equation $5x_1+21x_2+171x_3=268$ has three different non-negative solutions (representations): $(11,2,1)$, $(20,8,0)$ and $(41,3,0)$. However, $N=268$ is not the largest integer that has three solutions, but $N=289$ is the largest. This implies that for all $N>289$, the linear equation $5x_1+21x_2+171x_3=N$ has at least four solutions. We are interested in an explicit form of such a largest integer N.

More generally, for positive integers $a_1,\cdots ,a_n$ with $gcd(a_1,\cdots ,a_n)=1$ and N, we shall consider the linear equation

$$a_1{x}_1+\cdots +a_n{x}_n=N.$$

(2)

For a given non-negative integer p, the largest integer N, such that the linear Equation (2) has at most p different non-negative solutions (representations), is called the p-Frobenius number and denoted by $g_p(a_1,\cdots ,a_n)$ [5,6,7]. When $p=0$, 0-Frobenius number is called the Frobenius number, and such a problem is called the linear Diophantine problem of Frobenius. This problem, also known as the stamp exchange problem or the chicken McNuggets problem, has long fascinated many people in various ways. Much research has been done on this issue from various perspectives, but what we are interested in is the expression of closed formulas. Indeed, when $p=0$, for two variables, Sylvester obtained the explicit form $g(a_1,a_2)=g_0(a_1,a_2)=(a_1-1)(a_2-1)-1$. For three (or more) variables, however, no explicit form of $g(a_1,a_2,a_3)$ has been found, but the Frobenius number cannot be given by any set of closed formulas which can be reduced to a finite set of certain polynomials [8]. Only some special cases, explicit closed formulas have been found, including arithmetic, geometric-like, Fibonacci, Mersenne, and triangular (see [9,10,11,12,13,14,15] and references therein). When $p>0$, the situation is even harder. For two variables, it is known that $g_p(a_1,a_2)=(p+1)a_1{a}_2-a_1-a_2$. However, for three or more variables, no concrete example had been found. Most recently, we have finally succeeded in giving the p-Frobenius number as a closed-form expression of the triangular number triplet [5]. Results on p-Frobenius numbers for repunits [6] and Fibonacci triples [16] will also be available.

In this paper, we find closed forms of p-Frobenius number of the linear Equation (1). In detail, we give complete expressions of p-Frobenius numbers about p-Frobenius numbers for $p=0,1,2,3$, and for general p when $⌊(J_i-1)/J_k⌋\ge p$.

The Sylvester number (or genus) is as important as the Frobenius number. That is, the number of positive integers N such that the linear Equation (2) has at most p non-negative integer solutions is called the p-Sylvester number (or p-genus), and is denoted by $n_p(a_1,\cdots ,a_n)$. When $p=0$, 0-Sylvester number is the original Sylvester number. The concept of genus is very important in studying algebraic geometry, in particular, algebraic curves. Similarly to Frobenius numbers, Sylvester obtained the explicit form $n(a_1,a_2)=n_0(a_1,a_2)=(a_1-1)(a_2-1)/2$, but for three (or more) variables, no explicit form of $n(a_1,a_2,a_3)$ has been found. If $p>0$, the problem becomes even harder too.

In this paper, we also find closed forms of p-genus of the linear Equation (1). In order to compute both p-Frobenius numbers and p-genus, we analyze the geometrical structure of p-Apéry set. This framework is partly similar to that in Fibonacci triples [16], but complete descriptions require delicate case-by-case comparisons.

Throughout the paper, some known identities on Jacobsthal numbers are used. Two of the most important ones are $J_n=2J_{n-1}-{(-1)}^n$ (exercise 12 [1], [Chapter 44]) and $J_{i+k}=J_k{J}_{i+2}-4J_{k-2}{J}_i$ (exercise 26 in [1], [Chapter 44]).

2. Initial Results

For integers $i,k\ge 3$, set the integers r and ℓ by

$$J_i-1=r{J}_k+\ell ,0\le \ell \le J_k-1.$$

Theorem 1.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill & g_0(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\left\{\begin{array}{cc}(J_i-r{J}_k-1)J_{i+2}+r{J}_{i+k}-J_i\hfill & \mathit{if}r=0,\mathrm{or}r\ge 1\mathit{and}\hfill \\ \hfill & (J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i;\hfill \\ (J_k-1)J_{i+2}+(r-1)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i.\hfill \end{array}\right.\hfill \end{array}$$

(3)

Remark 1.

When $r=0$, that is, $k\ge i$, we have $g_0(J_i,J_{i+2},J_{i+k})=g_0(J_i,J_{i+2})$.

Theorem 2.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill g_1(J_i,J_{i+2},J_{i+k})& =(2J_i-1)J_{i+2}-J_i(k\ge i+2),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill g_1(J_i,J_{i+2},J_{2i+1})& =\left\{\begin{array}{cc}(J_{i+1}-1)J_{i+2}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ (2J_i-1)J_{i+2}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill g_1(J_i,J_{i+2},J_{2i})& =(J_i-1)J_{i+2}+J_{2i}-J_i.\hfill \end{array}$$

(6)

When $r=⌊(J_i-1)/J_k⌋\ge 1$, that is, $k\le i-1$ ($i\ge 4$), we have

$$\begin{array}{cc}\hfill & g_1(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\left\{\begin{array}{cc}(J_i-r{J}_k-1)J_{i+2}+(r+1)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i;\hfill \\ (J_k-1)J_{i+2}+r{J}_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i.\hfill \end{array}\right.\hfill \end{array}$$

(7)

Remark 2.

For example, when $k=i-1$, we obtain

$$\begin{array}{cc}\hfill g_1(J_i,J_{i+2},J_{2i-1})& =g_1(J_i,J_{i+2})-\left\{\begin{array}{cc}(J_6+J_2)\hfill & \mathit{if}i=3;\hfill \\ (J_{2i}+J_{i+2}+J_{i-1})\hfill & \mathit{if}i\ge 5\mathit{is}\mathit{odd};\hfill \\ (J_{2i}-J_{i-1})\hfill & \mathit{if}i\mathit{is}\mathit{even}.\hfill \end{array}\right.\hfill \end{array}$$

Concerning the relation with $g_1(J_i,J_{i+2})$, the identities (4), (5) and (6) can be written as

$$\begin{array}{cc}\hfill g_1(J_i,J_{i+2},J_{i+k})& =g_1(J_i,J_{i+2})(k\ge i+2),\hfill \\ \hfill g_1(J_i,J_{i+2},J_{2i+1})& =g_1(J_i,J_{i+2})-\left\{\begin{array}{cc}{J}_{i+2}\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ 0\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \\ \hfill g_1(J_i,J_{i+2},J_{2i})& =g_1(J_i,J_{i+2})-4J_i{J}_{i-2}.,\hfill \end{array}$$

respectively.

Theorem 3.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill g_2(J_i,J_{i+2},J_{i+k})& =(3J_i-1)J_{i+2}-J_i(k\ge i+2),\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill g_2(J_i,J_{i+2},J_{2i+1})& =\left\{\begin{array}{cc}(J_i-1)J_{i+2}+J_{2i+1}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ (J_i-2)J_{i+2}+J_{2i+1}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill g_2(J_i,J_{i+2},J_{2i})& =(2J_i-1)J_{i+2}-J_i,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill g_2(J_i,J_{i+2},J_{2i-1})& =(J_{i-1}-3)J_{i+2}+3J_{2i-1}-J_i\left(\mathit{if}i\mathit{is}\mathit{even}\right).\hfill \end{array}$$

(11)

When $r=⌊(J_i-1)/J_k⌋\ge 2$, that is, $k\le i-1$ ($i\ge 5$:odd) or $k\le i-2$ ($i\ge 6$:even), we have

$$\begin{array}{cc}\hfill & g_2(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\left\{\begin{array}{cc}(J_i-r{J}_k-1)J_{i+2}+(r+2)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i;\hfill \\ (J_k-1)J_{i+2}+(r+1)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i.\hfill \end{array}\right.\hfill \end{array}$$

(12)

Remark 3.

Concerning the relation with $g_2(J_i,J_{i+2})$, the identities (8), (9), (10) and (11) can be written as

$$\begin{array}{cc}\hfill g_2(J_i,J_{i+2},J_{i+k})& =g_2(J_i,J_{i+2})(k\ge i+2),\hfill \\ \hfill g_2(J_i,J_{i+2},J_{2i+1})& =g_2(J_i,J_{i+2})-\left\{\begin{array}{cc}(J_{i-1}{J}_{i+2}+J_{i-1}+J_{i+2})\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ (J_{i-1}{J}_{i+2}-J_{i-1})\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \\ \hfill g_2(J_i,J_{i+2},J_{2i})& =g_2(J_i,J_{i+2})-J_i{J}_{i+2},\hfill \\ \hfill g_2(J_i,J_{i+2},J_{2i-1})& =g_2(J_i,J_{i+2})-11J_{i-1}{J}_i\left(\mathit{if}i\mathit{is}\mathit{even}\right),\hfill \end{array}$$

respectively.

Theorem 4.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill g_3(J_i,J_{i+2},J_{i+k})& =(4J_i-1)J_{i+2}-J_i(k\ge i+2),\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill g_3(J_i,J_{i+2},J_{2i+1})& =\left\{\begin{array}{cc}(3J_i-2)J_{i+2}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ (3J_i-1)J_{i+2}-J_i\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill g_3(J_i,J_{i+2},J_{2i})& =\left\{\begin{array}{cc}(3J_i-1)J_{i+2}-(J_{i+1}+4)J_i\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ (3J_i-1)J_{i+2}-(J_{i+1}-2)J_i\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \end{array}$$

(15)

When $r=⌊(J_i-1)/J_k⌋\ge 3$, that is, $k\le i-2$ ($i\ge 5$), we have

$$\begin{array}{cc}\hfill & g_3(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\left\{\begin{array}{cc}(J_i-r{J}_k-1)J_{i+2}+(r+3)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i;\hfill \\ (J_k-1)J_{i+2}+(r+2)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i.\hfill \end{array}\right.\hfill \end{array}$$

(16)

Remark 4.

Concerning the relation with $g_3(J_i,J_{i+2})$, the identities (13), (14) and (15) can be written as

$$\begin{array}{cc}\hfill g_3(J_i,J_{i+2},J_{i+k})& =g_3(J_i,J_{i+2})(k\ge i+2),\hfill \\ \hfill g_3(J_i,J_{i+2},J_{2i+1})& =g_3(J_i,J_{i+2})-\left\{\begin{array}{cc}(J_i+1)J_{i+2}\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ J_i{J}_{i+2}\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \\ \hfill g_3(J_i,J_{i+2},J_{2i})& =g_3(J_i,J_{i+2})-\left\{\begin{array}{cc}{J}_i(J_{i+2}+J_{i+1}+3)\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ J_i(J_{i+2}+J_{i+1}-3)\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \end{array}$$

respectively.

3. Preliminaries

Without loss of generality, we assume that $a_1={min}_{1\le j\le k}\left\{a_j\right\}$. For each $0\le i\le a_1-1$, we introduce the positive integer $m_i^{\left(p\right)}$ congruent to i modulo $a_1$, such that the number of representations of $m_i^{\left(p\right)}$ is bigger than or equal to $p+1$ and that of $m_i^{\left(p\right)}-a_1$ is less than or equal to p. Note that $m_0^{\left(0\right)}$ is defined to be 0. The set

$${\mathrm{Ap}}_p\left(A\right)={\mathrm{Ap}}_p(a_1,a_2,\cdots ,a_k)=\{m_0^{\left(p\right)},m_1^{\left(p\right)},\cdots ,m_{a_1-1}^{\left(p\right)}\}$$

is called the p-Apéry set of $A=\{a_1,a_2,\cdots ,a_k\}$ for a non-negative integer p (see, [17]), which is congruent, modulo $a_1$, to the set

$$\{0,1,\cdots ,a_1-1\}.$$

When $p=0$, 0-Apéry set is the original Apéry set.

If one can know the structure of the elements of p-Apéry set, we can obtain the p-Frobenius numbers and p-gaps (p-Sylvester numbers) by the following formulas. A more general form can be seen in [7].

Lemma 1.

Let k and p be integers with $k\ge 2$ and $p\ge 0$.

Assume that $gcd(a_1,a_2,\cdots ,a_k)=1$. We have

$$\begin{array}{cc}\hfill g_p(a_1,a_2,\cdots ,a_k)& =\underset{0\le j\le a_1-1}{max}{m}_j^{\left(p\right)}-a_1,\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill n_p(a_1,a_2,\cdots ,a_k)& =\frac{1}{a_1}\sum _{j=0}^{a_1-1}{m}_j^{\left(p\right)}-\frac{a_1-1}{2}.\hfill \end{array}$$

(18)

By using GAP package numericalsgps [18] (the authors thank Professor Pedro A. García-Sánchez for his kind instruction), p-Frobenius numbers of Jacobsthal triples can be obtained systematically.

• gap> jacob:= function(n)
• > if n=0 then
• > return 0;
• > elif n in [1, 2] then
• > return 1;
• > else
• > return jacob(n-1) + 2*jacob(n-2);
• > fi;
• > end;
• function( n ) … end
• gap> jacob(15);
• 10923
• gap> Set([3..10],i->FrobeniusNumber(DenumerantIdeal
• (NumericalSemigroup (jacob(i),jacob(i+2),jacob(2*i-1)),3)));
• [118, 145, 718, 2960, 12430, 50064, 202638, 811920 ]

For example,

$${\left\{g_3(J_i,J_{i+2},J_{2i-1})\right\}}_{i=3}^{10}=118,145,718,2960,12430,50064,202638,811920$$

(It is quite immediate to obtain the values until 50064. However, even by using GAP, it takes several hours to obtain 811920.)

We can have the 1-Apéry set and the representations of each element. For example, for $(J_5,J_7,J_9)=(11,43,171)$, we can obtain the 1-Apéry set ${\mathrm{Ap}}_1(11,43,171)=\{385,386,387,344,301,258,215,513,514,471,428\}$, and each element has exactly two representations. For instance, $385=1·43+2·171=35·11$.

• gap> s:=NumericalSemigroup(11,43,171);
• gap> s1:=DenumerantIdeal(s,1);
• gap> AperyList(s1,11);
• [385, 386, 387, 344, 301, 258, 215, 513, 514, 471, 428 ]
• gap> FactorizationsElementListWRTNumericalSemigroup(AperyList(s1,11)
• ,s);
• [[[0, 1, 2 ], [35, 0, 0 ]], [[0, 5, 1 ], [4, 0, 2 ]], [[0, 9, 0 ], [4, 4, 1 ]],
• [[0, 8, 0 ], [4, 3, 1 ]], [[0, 7, 0 ], [4, 2, 1 ]], [[0, 6, 0 ], [4, 1, 1 ]],
• [[0, 5, 0 ], [4, 0, 1 ]], [[0, 0, 3 ], [31, 4, 0 ]], [[0, 4, 2 ], [35, 3, 0 ]],
• [[0, 3, 2 ], [35, 2, 0 ]], [[0, 2, 2 ], [35, 1, 0 ]]]

4. Proof of the Main Results

Put the linear representation

$$\begin{array}{cc}\hfill t_{x,y}:& =x{J}_{i+2}+y{J}_{i+k}\hfill \\ \hfill & =(x+y{J}_k)J_{i+2}-4y{J}_{k-2}{J}_i(x,y\ge 0).\hfill \end{array}$$

Given two Jacobsthal numbers $J_i$ and $J_k$, integers r and ℓ are determined uniquely by

$$J_i-1=r{J}_k+\ell \mathrm{with}0\le \ell \le J_k-1.$$

Proof of Theorem 1.

Since $gcd(J_i,J_{i+2})=1$ ($i\ge 3$), for $0\le x\le J_k-1$ and $0\le y\le r-1$ or $0\le x\le \ell$ and $y=r$, we have each element of $t_{x,y}$ belongs to ${\mathrm{Ap}}_0(J_i,J_{i+2},J_{i+k})$. See Chart 1. Namely, $t_{x,y}\equiv t_{x^{\prime },y^{\prime }}\left(\mathrm{mod}{J}_i\right)$ implies that $x=x^{\prime }$ and $y=y^{\prime }$. When $r=0$, the largest possible element is $t_{\ell ,r}$. When $r\ge 1$, the largest possible element is $t_{\ell ,r}$ or $t_{J_k-1,r-1}$, and $t_{\ell ,r}>t_{J_k-1,r-1}$ is equivalent to $(J_i-r{J}_k)J_{i+2}>4J_{k-2}{J}_i$. Hence, by (17), we obtain the desired result. □

Proof of Theorem 2.

First, assume that $r=⌊(J_i-1)/J_k⌋\ge 1$. Since there are three congruence relations

$$\begin{array}{cc}\hfill t_{x,y}& \equiv t_{J_k+x,y-1}\left(\mathrm{mod}{J}_i\right)(0\le x\le J_k-1,1\le y\le r-1;\hfill \\ \hfill & 0\le x\le \ell ,y=r),\hfill \\ \hfill t_{x,0}& \equiv t_{\ell +1+x,r}\left(\mathrm{mod}{J}_i\right)(0\le x\le J_k-\ell -2),\hfill \\ \hfill t_{J_k-\ell -1+x,0}& \equiv t_{x,r+1}\left(\mathrm{mod}{J}_i\right)(0\le x\le \ell ),\hfill \end{array}$$

we can obtain the 1-Apéry set ${\mathrm{Ap}}_1\left(A\right)$ from the 0-Apéry set ${\mathrm{Ap}}_0\left(A\right)$, where $A=\{J_i,J_{i+2},J_{i+k}\}$. By $J_{i+k}=J_k{J}_{i+2}-4J_{k-2}{J}_i$, we can see that each element of ${\mathrm{Ap}}_1\left(A\right)$ has exactly two different expressions in terms of $J_i,J_{i+2},J_{i+k}$:

$$\begin{array}{cc}\hfill t_{J_k+x,y-1}& =(J_k+x)J_{i+2}+(y-1)J_{i+k}\hfill \\ \hfill & =4J_{k-2}{J}_i+x{J}_{i+2}+y{J}_{i+k},\hfill \\ \hfill t_{\ell +1+x,r}& =(\ell +1+x)J_{i+2}+r{J}_{i+k}\hfill \\ \hfill & =(J_{i+2}-4r{J}_{k-2})J_i+x{J}_{i+2},\hfill \\ \hfill t_{x,r+1}& =x{J}_{i+2}+(r+1)J_{i+k}\hfill \\ \hfill & =\left(J_{i+2}-4(r+1)J_{k-2}\right)J_i+(x+J_k-\ell -1)J_{i+2}.\hfill \end{array}$$

Concerning the largest value in the set ${\mathrm{Ap}}_1\left(A\right)$, as seen in Chart 2, there are four candidates:

$$t_{\ell ,r+1},t_{J_k-1,r},t_{J_k+\ell ,r-1},t_{2J_k-1,r-2}.$$

Since $J_{i+2}/J_i\ge 3$ ($i\ge 1$), we can know that

$$t_{\ell ,r+1}>t_{J_k+\ell ,r-1}\mathrm{and}{t}_{J_k-1,r}>t_{2J_k-1,r-2}.$$

Therefore, it is enough to compare $t_{\ell ,r+1}$ and $t_{J_k-1,r}$. Since $t_{\ell ,r+1}>t_{J_k-1,r}$ is equivalent to $(J_i-r{J}_k)J_{i+2}>4J_{k-2}{J}_i$, we have the identity (7).

The identities (4), (5) and (6) are yielded from the case $r=0$. In this case, there are two different patterns for the arrangement of the elements of 1-Apéry set: $2\ell +1\le J_k-1$ or $2\ell +1>J_k-1$.

Case 1.

When $2\ell +1\le J_k-1$, that is, $2J_i\le J_k$, all the elements of 0-Apéry set and 1-Apéry set are in a unique line, as shown in Chart 3. There is the relation between each element in 0-Apéry set and 1-Apéry set as $t_{j,0}\equiv t_{\ell +j+1,0}\left(\mathrm{mod}{J}_i\right)$. Each element in 1-Apéry set has exactly two representations because for $0\le j\le \ell$

$$\begin{array}{cc}\hfill t_{j,0}& =j{J}_{i+2}\equiv (J_i+j)J_{i+2}\hfill \\ \hfill & =(\ell +j+1)J_{i+2}=t_{\ell +j+1,0}\left(\mathrm{mod}{J}_i\right).\hfill \end{array}$$

Hence, the largest element of 1-Apéry set is

$$t_{2\ell +1,0}=(2\ell +1)J_{i+2}=(2J_i-1)J_{i+2}.$$

Notice that the condition $2\ell +1\le J_k-1$ is valid when $k\ge i+2$ or $k=i+1$ and i is even. By (17), we obtain the identity (4), and the identity (5) when i is even.

Case 2.

When $2\ell +1>J_k-1$, that is, $2J_i>J_k$, some elements of 1-Apéry set come to the second line, as shown in Chart 4. There are two corresponding relations between each element in 0-Apéry set and 1-Apéry set: $t_{j,0}\equiv t_{\ell +j+1,0}$ ($0\le j\le J_k-\ell -2$) and $t_{J_k-\ell +j-1,0}\equiv t_{j,1}\left(\mathrm{mod}{J}_j\right)$ ($0\le j\le 2\ell +1-J_k$). Each element in 1-Apéry set has exactly two representations because for $0\le j\le J_k-\ell -2$

$$\begin{array}{cc}\hfill t_{j,0}& =j{J}_{i+2}\equiv (J_i+j)J_{i+2}\hfill \\ \hfill & =(\ell +1+j)J_{i+2}=t_{\ell +1+j,0}\left(\mathrm{mod}{J}_i\right)\hfill \end{array}$$

and for $0\le j\le 2\ell +1-J_k$

$$\begin{array}{cc}\hfill t_{J_k-\ell -1+j,0}& =(J_k-\ell -1+j)J_{i+2}\hfill \\ \hfill & \equiv (j+J_k)J_{i+2}-4J_{k-2}{J}_i=j{J}_{i+2}+J_{i+k}=t_{j,1}\left(\mathrm{mod}{J}_i\right).\hfill \end{array}$$

Hence, concerning the largest element of 1-Apéry set, there are two possibilities: $t_{J_k-1,0}$ and $t_{2\ell +1-J_k,1}$. Since $r=0$ implies that $k\ge i$, the condition $2J_i>J_k$ is valid only when $k=i$ or $k=i+1$ and i is odd. When $k=i$, only possibility is $t_{2\ell +1-J_k,1}=(J_i-1)J_{i+2}+J_{2i}$, yielding the identity (6). When $k=i+1$ (and i is odd), by $t_{J_k-1,0}>t_{2\ell +1-J_k,1}$, we obtain the identity (5) when i is odd. □

Proof of Theorem 3.

When $p=2$, the elements of 2-Apéry set are determined from those of 1-Apéry set, by using the third block too (see Chart 5). Except for the elements in the top row, each element in the 1-Apéry set corresponds to the element in the 2-Apéry set with the same residue modulo $J_i$ so that it shifts up one step and moves to the right block. Only the elements in the 1-Apéry in the top row correspond to the elements of the 2-Apéry set with the same residue so that they are arranged in order so as to fill the gap at the bottom of the leftmost block (first block).

Case 1.

When $r\ge 2$, the elements of 2-Apéry set are placed below the first block, below the second block, and throughout the third block, as shown in Chart 6. Note that when $r=2$, the third block consists of only one line with $J_k$ or less elements.

Each element in the lower two lines of the first block (or just one line consisting of $J_k$ elements) can be represented in only three ways in terms of $J_i$, $J_{i+2}$ and $J_{i+k}$, respectively, and is congruent to an element in the top row of the second block modulo $J_i$. Each element in the lower two lines of the second block (or just one line consisting of $J_k$ elements) can be represented in only three ways in terms of $J_i$, $J_{i+2}$ and $J_{i+k}$, respectively, and is congruent to an element in the second row from the bottom of the first block modulo $J_i$. Every element of the third block has also only three ways of being represented in terms of $J_i$, $J_{i+2}$ and $J_{i+k}$, and there is a congruence relation with each element (except the first row) of the second block shifted up by one step.

Concerning the largest element of ${\mathrm{Ap}}_2(J_i,J_{i+2},J_{i+k})$, there are six candidates:

$$t_{\ell ,r+2},t_{J_k-1,r+1},t_{J_k+\ell ,r},t_{2J_k-1,r-1},t_{2J_k+\ell ,r-2},t_{3J_k-1,r-3},.$$

Since $J_{i+2}\ge 3J_i$ ($j\ge 1$), $2J_{i+k}>J_k{J}_{i+2}$. Hence, we obtain

$$\begin{array}{cc}\hfill t_{\ell ,r+2}& >t_{J_k+\ell ,r}>t_{2J_k+\ell ,r-2},\hfill \\ \hfill t_{J_k-1,r+1}& >t_{2J_k-1,r-1}>t_{3J_k-1,r-3}].\hfill \end{array}$$

Therefore, the largest value must be $t_{\ell ,r+2}$ or $t_{J_k-1,r+1}$. If $(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i$, the largest value in ${\mathrm{Ap}}_2\left(A\right)$ is $t_{\ell ,r+2}=(J_i-r{J}_k-1)J_{i+2}+(r+2)J_{i+k}$. If $(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i$, the largest value in ${\mathrm{Ap}}_2\left(A\right)$ is $t_{J_k-1,r+1}=(J_k-1)J_{i+2}+(r+1)J_{i+k}$. By (17), we obtain the identity (12).

Case 2.

When $r=1$, there are two patterns depending on $J_k\ge 2\ell +2$ or $J_k<2\ell +2$.

If $J_k\ge 2\ell +2$ then $3J_k\ge 2J_i$, so $k\ge i$.

However, by $r=1$, $k\le i-1$. So, this case does not exist. If $J_k<2\ell +2$, by $r=1$, the only possibility is $k=i-1$ and i is even. As shown in Chart 7, there are four candidates for the largest value of ${\mathrm{Ap}}_2\left(A\right)$:

$$\begin{array}{cc}\hfill t_{2\ell +1-J_k,3}& =(2J_i-3J_{i-1}-1)J_{i+2}+3J_{2i-1},,\hfill \\ \hfill t_{J_k-1,2}& =(J_{i-1}-1)J_{i+2}+2J_{2i-1},\hfill \\ \hfill t_{J_k+\ell ,1}& =(J_i-1)J_{i+2}+J_{2i-1},\hfill \\ \hfill t_{2J_k-1,0}& =(2J_{i-1}-1)J_{i+2}.\hfill \end{array}$$

Since $2J_{i-1}-J_i=1$ when i is even, by $J_{2i-1}>2J_{i-2}{J}_{i+2}\ge 2J_{i+2}$ we have

$$t_{2\ell +1-J_k,3}>t_{J_k-1,2}>t_{J_k+\ell ,1}>t_{2J_k-1,0}.$$

By (17), we can obtain the identity (11).

Case 3.

When $r=0$, there are three patterns, depending on $J_k\ge 3\ell +3$, $2\ell +2\le J_k\le 3\ell +2$ or $J_k\le 2\ell +1$. If $J_k\ge 3\ell +3=3J_i$, we see that $k\ge i+2$. As shown in Chart 8, the largest value of ${\mathrm{Ap}}_2\left(A\right)$ is $t_{3\ell +2,0}=(3J_i-1)J_{i+2}$. By (17), we obtain the identity (8).

If $2\ell +2\le J_k\le 3\ell +2$, that is, $2J_i\le J_k\le 3J_i-1$, together with $J_i-1<J_k$, only the possibility is $k=i+1$ and i is even. As shown in Chart 9, the largest value of ${\mathrm{Ap}}_2\left(A\right)$ is $t_{3\ell +2-J_k,1}$ or $t_{J_k-1,0}$. Since $J_{2i+1}>(J_i+2)J_{i+2}=(2J_{i+1}-3J_i)J_{i+2}$ for even i, we have $t_{3\ell +2-J_k,1}>t_{J_k-1,0}$. By (17), we obtain the identity (9) when i is even.

If $J_k\le 2\ell +1$, by $J_i-1<J_k\le 2J_i-1$, we see that $k=i$ or $k=i+1$ and i is odd. As shown in Chart 10, the largest value of ${\mathrm{Ap}}_2\left(A\right)$ is $t_{\ell ,1}$ or $t_{2\ell +1,0}$. When $k=i+1$ (and i is odd), we have $t_{\ell ,1}>t_{2\ell +1,0}$. When $k=i$, we have $t_{\ell ,1}<t_{2\ell +1,0}$. By (17), we obtain the identity (9) when i is odd, and the identity (10), respectively. □

Proof of Theorem 4.

When $p=3$, the structure will be systematically analyzed if $r\ge 3$. On the contrary, in the case where $r<3$, the description is not simple at all because the delicate discussion of each case is required. Here, we outline the case of $r\ge 3$ for generalization in a later section.

In Chart 11, ⓝ denotes the area of elements in ${\mathrm{Ap}}_n\left(A\right)$. Here, each $m_j^{\left(n\right)}$, satisfying $m_j^{\left(n\right)}\equiv j\left(\mathrm{mod}{J}_i\right)$ ($0\le j\le J_i-1$), can be expressed in at least $n+1$ ways but $m_j^{\left(n\right)}-J_i$ in at most n ways. Each element of ${\mathrm{Ap}}_3\left(A\right)$ existing in the second block to the fourth block corresponds to each element having the same residue of ${\mathrm{Ap}}_2\left(A\right)$ existing in the block immediately to the left thereof in a form of shifting up one step. The $J_k$ elements of ${\mathrm{Ap}}_3\left(A\right)$ existing over two rows (or one row) at the bottom of the first block correspond to the $J_k$ elements with the same residue of ${\mathrm{Ap}}_2\left(A\right)$ at the top of the third block. Therefore, since all the elements in ${\mathrm{Ap}}_2\left(A\right)$ form a complete remainder system, so is ${\mathrm{Ap}}_3\left(A\right)$. We omit the details, but it can be seen that all the elements of ${\mathrm{Ap}}_3\left(A\right)$ have exactly four ways of expressing in terms of $J_i$, $J_{i+2}$, and $J_{i+k}$. Within each region of ${\mathrm{Ap}}_3\left(A\right)$, the rightmost (lower right) element is the largest, so by comparing those elements, the maximum element of ${\mathrm{Ap}}_3\left(A\right)$ can be found.

Eventually, the largest element of ${\mathrm{Ap}}_3\left(A\right)$ is from the first block, that is, $t_{\ell ,r+3}=(J_i-r{J}_k-1)J_{i+2}+(r+3)J_{i+k}$ or $t_{J_k-1,r+2}=(J_k-1)J_{i+2}+(r+2)J_{i+k}$. Hence, if $(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i$, then $g_3(J_i,J_{i+2},J_{i+k})=t_{\ell ,r+3}-J_i=(J_i-r{J}_k-1)J_{i+2}+(r+3)J_{i+k}-J_i$. If $(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i$, then $g_3(J_i,J_{i+2},J_{i+k})=t_{J_k-1,r+2}-J_i=(J_k-1)J_{i+2}+(r+2)J_{i+k}-J_i$. Notice that $r\ge 3$ implies that $k\le i-2$ with $i\ge 5$. □

5. General p Case

Let p be a given non-negative integer. When $r\ge p$, by repeating the same process as those in previous theorems, we can obtain an explicit formula of p-Frobenius number.

Theorem 5.

Let $i,k\ge 3$ and p be a non-negative integer. When $r=⌊(J_i-1)/J_k⌋\ge p$ with $(r,p)\ne (0,0)$, we have

$$\begin{array}{cc}\hfill & g_p(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\left\{\begin{array}{cc}(J_i-r{J}_k-1)J_{i+2}+(r+p)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}\ge 4J_{k-2}{J}_i;\hfill \\ (J_k-1)J_{i+2}+(r+p-1)J_{i+k}-J_i\hfill & \mathit{if}(J_i-r{J}_k)J_{i+2}<4J_{k-2}{J}_i.\hfill \end{array}\right.\hfill \end{array}$$

(19)

However, when $r<p$, the situation is not simple at all. We do need the case-by-case discussion. To see this depends on how regularly the elements of the p-Apéry set continues to appear. The elements of the 0-Apéry set represented by Chart 2 are continually arranged regularly to those of the 1-Apéry set, the 2-Apéry set, and until the r-Apéry set. However, when it reaches the time of the (p)-Apéry set when $p=r+1$, the elements arranged at the lower left are insufficient, and the regularity shifts. That is, the positional relationship of the elements giving the largest value suddenly changes. Hereinafter, this deviation increases as $p=r+2$, $p=r+3$, ⋯, and the regularity is greatly impaired.

Chart 12 shows the arrangement of the p-Apéry sets ($p=0,1,\cdots ,5$) when $r=5$. One can see that there will be a deficiency in the arrangement of the lower left for 6-Apéry set.

6. p-Genus

In order to obtain an explicit form of p-genus (p-Sylvester number), we shall use the Formula (18). That is, in each case, it is necessary to find the sum of the elements within the areas of p-Apéry set. Calculating the sum is a bit cumbersome, but there is no need to compare candidates as we would when finding the largest value. Therefore, the calculation method is similar to that of Fibonacci [16].

6.1. The Case $p=0$

By Chart 1, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{J_i-1}{m}_j^{\left(0\right)}& =\sum _{x=0}^{J_k-1}\sum _{y=0}^{r-1}{t}_{x,y}+\sum _{x=0}^{\ell }{t}_{x,r}\hfill \\ \hfill & =\frac{r(J_k-1)J_k}{2}{J}_{i+2}+\frac{J_k(r-1)r}{2}{J}_{i+k}+\frac{\ell (\ell +1)}{2}{J}_{i+2}+(\ell +1)r{J}_{i+k}\hfill \\ \hfill & =\frac{J_i^2+r(r+1)J_k^2-J_i(2r{J}_k+1)}{2}{J}_{i+2}\hfill \\ \hfill & +\frac{2J_i-(r+1)J_k}{2}r(J_{i+2}{J}_k-4J_i{J}_{k-2})\hfill \\ \hfill & =\frac{J_i(J_i-1)J_{i+2}}{2}-2r{J}_i\left(2J_i-(r+1)J_k\right)J_{k-2}.\hfill \end{array}$$

By (18), we have

$$\begin{array}{cc}\hfill n_0(J_i,J_{i+2},J_{i+k})& =\frac{1}{J_i}\sum _{j=0}^{J_i-1}{m}_j^{\left(0\right)}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{(J_i-1)J_{i+2}}{2}-2r\left(2J_i-(r+1)J_k\right)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{(J_i-1)(J_{i+2}-1)}{2}-2r{J}_i\left(2J_i-(r+1)J_k\right)J_{k-2}.\hfill \end{array}$$

Theorem 6.

We have

$$n_0(J_i,J_{i+2},J_{i+k})=\frac{(J_i-1)(J_{i+2}-1)}{2}-2r{J}_i\left(2J_i-(r+1)J_k\right)J_{k-2}.$$

Remark 5.

For Fibonacci triples [19], we have

$$n_0(F_i,F_{i+2},F_{i+k})=\frac{(F_i-1)(F_{i+2}-1)}{2}-\frac{r\left(2F_i-(r+1)F_k\right)F_{k-2}}{2}.$$

6.2. The Case $p=1$

When $r\ge 1$, by Chart 2, we have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{J_i-1}{m}_j^{\left(1\right)}=\sum _{x=J_k}^{2J_k-1}\sum _{y=0}^{r-2}{t}_{x,y}+\sum _{x=J_k}^{J_k+\ell }{t}_{x,r-1}+\sum _{x=\ell +1}^{J_k-1}{t}_{x,r}+\sum _{x=0}^{\ell }{t}_{x,r+1}\hfill \\ \hfill & =\left(\frac{(2J_k-1)\left(2J_k\right)}{2}-\frac{(J_k-1)J_k}{2}\right)(r-1)J_{i+2}+\frac{(r-2)(r-1)}{2}{J}_k{J}_{i+k}\hfill \\ \hfill & +\left(\frac{(J_k+\ell )(J_k+\ell +1)}{2}-\frac{(J_k-1)J_k}{2}\right)J_{i+2}+(\ell +1)(r-1)J_{i+k}\hfill \\ \hfill & +\left(\frac{(J_k-1)\left(J_k\right)}{2}-\frac{\ell (\ell +1)}{2}\right)J_{i+2}+(J_k-\ell -1)r{J}_{i+k}\hfill \\ \hfill & +\frac{\ell (\ell +1)}{2}{J}_{i+2}+(\ell +1)(r+1)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}\left((r-1)(3J_k-1)J_k+(J_i-1-(r-1)J_k)(J_i-(r-1)J_k)\right)J_{i+2}\hfill \\ \hfill & +\left(r{J}_i-\frac{(r-1)(r+2)}{2}{J}_k\right)(J_k{J}_{i+2}-4J_{k-2}{J}_i)\hfill \\ \hfill & =\frac{1}{2}(J_i+2J_k-1)J_i{J}_{i+2}-4\left(r{J}_i-\frac{(r-1)(r+2)}{2}{J}_k\right)J_{k-2}{J}_i.\hfill \end{array}$$

Hence, by (18) we have

$$\begin{array}{cc}\hfill & n_1(J_i,J_{i+2},J_{i+k})=\frac{1}{J_i}\sum _{j=0}^{J_i-1}{m}_j^{\left(0\right)}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}(J_i+2J_k-1)J_{i+2}-2\left(2r{J}_i-(r-1)(r+2)J_k\right)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}\left((J_i+2J_k-1)J_{i+2}-J_i+1\right)-4J_k{J}_{k+2}-2r\left(2J_i-(r+1)J_k\right)J_{k-2}.\hfill \end{array}$$

When $r=0$ and $2\ell +1\le J_k-1$, that is, $k\ge i+2$ or $k=i+1$ and i is even, by using Chart 3, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{J_i-1}{m}_j^{\left(1\right)}& =\sum _{x=\ell +1}^{2\ell +1}{t}_{x,0}\hfill \\ \hfill & =\left(\frac{(2\ell +1)(2\ell +2)}{2}-\frac{\ell (\ell +1)}{2}\right)J_{i+2}\hfill \\ \hfill & =\frac{(3J_i-1)J_i{J}_{i+2}}{2}.\hfill \end{array}$$

Hence, by (18), we have

$$\begin{array}{cc}\hfill n_1(J_i,J_{i+2},J_{i+k})& =\frac{(3J_i-1)J_{i+2}}{2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1),\hfill \end{array}$$

When $r=0$ and $2\ell +1\ge J_k$, that is, $k=i$ or $k=i+1$ and i is odd, by Chart 4, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{J_i-1}{m}_j^{\left(1\right)}& =\sum _{x=\ell +1}^{J_k-1}{t}_{x,0}+\sum _{x=0}^{2\ell +1-J_k}{t}_{x,1}\hfill \\ \hfill & =\left(\frac{(J_k-1)J_k}{2}-\frac{\ell (\ell +1)}{2}\right)J_{i+2}\hfill \\ \hfill & +\frac{(2\ell +1-J_k)(2\ell +2-J_k)}{2}{J}_{i+2}+(2\ell +2-J_k)J_{i+k}\hfill \\ \hfill & =\left(\frac{(3J_i-1)J_i}{2}-2J_i{J}_k+J_k^2\right)J_{i+2}+(2J_i-J_k)J_{i+k}\hfill \\ \hfill & =\left(\frac{(3J_i-1)J_{i+2}}{2}-4(2J_i-J_k)J_{k-2}\right)J_i.\hfill \end{array}$$

Hence, by (18), we have

$$\begin{array}{cc}\hfill n_1(J_i,J_{i+2},J_{i+k})& =\frac{(3J_i-1)J_{i+2}}{2}-4(2J_i-J_k)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1)-4(2J_i-J_k)J_{k-2}.\hfill \end{array}$$

Theorem 7.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill & n_1(J_i,J_{i+2},J_{i+k})=\frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1)(k\ge i+2),\hfill \\ \hfill & n_1(J_i,J_{i+2},J_{2i+1})\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1)-4(2J_i-J_{i+1})J_{i-1}\hfill & i\mathit{is}\mathit{odd};\hfill \\ \frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1)\hfill & i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \\ \hfill & n_1(J_i,J_{i+2},J_{2i})=\frac{1}{2}(3J_i{J}_{i+2}-J_i-J_{i+2}+1)-4J_i{J}_{i-2}.\hfill \end{array}$$

When $r=⌊(J_i-1)/J_k⌋\ge 1$, that is, $k\le i-1$ ($i\ge 4$), we have

$$\begin{array}{cc}\hfill & n_1(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\frac{(J_i+2J_k-1)J_{i+2}-J_i+1)}{2}-2\left(2r{J}_i-(r+2)(r-1)J_k\right)J_{k-2}.\hfill \end{array}$$

6.3. The Case $p=2$

When $r\ge 2$, that is, $k\le i-1$ ($i\ge 5$) and i is odd or $k\le i-2$ ($i\ge 6$) and i is even, by Chart 6, we have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{J_i-1}{m}_j^{\left(2\right)}\hfill \\ \hfill & =\sum _{x=2J_k}^{3J_k-1}\sum _{y=0}^{r-3}{t}_{x,y}+\sum _{x=2J_k}^{2J_k+\ell }{t}_{x,r-2}+\sum _{x=J_k+\ell +1}^{2J_k-1}{t}_{x,r-1}+\sum _{x=J_k}^{J_k+\ell }{t}_{x,r}\hfill \\ \hfill & +\sum _{x=\ell +1}^{J_k-1}{t}_{x,r+1}+\sum _{x=0}^{\ell }{t}_{x,r+2}\hfill \\ \hfill & =(r-2)\left(\frac{(3J_k-1)\left(3J_k\right)}{2}-\frac{(2J_k-1)\left(2J_k\right)}{2}\right)J_{i+2}+\frac{(r-3)(r-2)}{2}{J}_k{J}_{i+k}\hfill \\ \hfill & +\left(\frac{(2J_k+\ell )(2J_k+\ell +1)}{2}-\frac{(2J_k-1)\left(2J_k\right)}{2}\right)J_{i+2}+(\ell +1)(r-2)J_{i+k}\hfill \\ \hfill & +\left(\frac{(2J_k-1)\left(2J_k\right)}{2}-\frac{(J_k+\ell )(J_k+\ell +1)}{2}\right)J_{i+2}+(J_k-\ell -1)(r-1)J_{i+k}\hfill \\ \hfill & +\left(\frac{(J_k+\ell )(J_k+\ell +1)}{2}-\frac{(J_k-1)J_k}{2}\right)J_{i+2}+(\ell +1)r{J}_{i+k}\hfill \\ \hfill & +\left(\frac{(J_k-1)J_k}{2}-\frac{\ell (\ell +1)}{2}\right)J_{i+2}+(J_k-\ell -1)(r+1)J_{i+k}\hfill \\ \hfill & +\frac{\ell (\ell +1)}{2}{J}_{i+2}+(\ell +1)(r+2)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}(J_i+4J_k-1)J_i{J}_{i+2}-2\left(2r{J}_i-(r+3)(r-2)J_k\right)J_i{J}_{k-2}.\hfill \end{array}$$

Hence, by (18), we have

$$\begin{array}{cc}\hfill & n_2(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\frac{1}{2}(J_i+4J_k-1)J_{i+2}-2\left(2r{J}_i-(r+3)(r-2)J_k\right)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}\left((J_i+4J_k-1)J_{i+2}-J_i+1\right)-2\left(2r{J}_i-(r+3)(r-2)J_k\right)J_{k-2}.\hfill \end{array}$$

When $r=1$, that is, $k=i-1$ and i is even, by Chart 7, we have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{J_i-1}{m}_j^{\left(2\right)}\hfill \\ \hfill & =\sum _{x=J_k+\ell +1}^{2J_k-1}{t}_{x,0}+\sum _{x=J_k}^{J_k+\ell }{t}_{x,1}+\sum _{x=\ell +1}^{J_k-1}{t}_{x,2}+\sum _{x=0}^{2\ell +1-J_k}{t}_{x,3}\hfill \\ \hfill & =\left(\frac{(2J_k-1)\left(2J_k\right)}{2}-\frac{(J_k+\ell )(J_k+\ell +1)}{2}\right)J_{i+2}\hfill \\ \hfill & +\left(\frac{(J_k+\ell )(J_k+\ell +1)}{2}-\frac{(J_k-1)J_k}{2}\right)J_{i+2}+(\ell +1)J_{i+k}\hfill \\ \hfill & +\left(\frac{(J_k-1)J_k}{2}-\frac{\ell (\ell +1)}{2}\right)J_{i+2}+2(J_k-\ell -1)J_{i+k}\hfill \\ \hfill & +\frac{(2\ell +1-J_k)(2\ell +2-J_k)}{2}{J}_{i+2}+3(2\ell +2-J_k)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}(3J_i-1)J_i{J}_{i+2}-4(5J_i-6J_k)J_i{J}_{k-2}.\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}\hfill n_2(J_i,J_{i+2},J_{i+k})& =\frac{1}{2}(3J_i-1)J_{i+2}-4(5J_i-6J_k)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}\left(3J_i{J}_{i+2}-J_i-J_{i+2}+1\right)-4(5J_i-6J_k)J_{k-2}.\hfill \end{array}$$

When $r=0$ and $J_k\ge 3\ell +3$, that is, $k\ge i+2$, by Chart 8, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{J_i-1}{m}_j^{\left(2\right)}& =\sum _{x=2\ell +2}^{3\ell +2}{t}_{x,0}\hfill \\ \hfill & =\left(\frac{(3\ell +2)(3\ell +3)}{2}-\frac{(2\ell +1)(2\ell +2)}{2}\right)J_{i+2}\hfill \\ \hfill & =\frac{1}{2}(5J_i-1)J_i{J}_{i+2}.\hfill \end{array}$$

Hence, by (18), we have

$$\begin{array}{cc}\hfill n_2(J_i,J_{i+2},J_{i+k})& =\frac{1}{2}(5J_i-1)J_{i+2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}(5J_i{J}_{i+2}-J_i-J_{i+2}+1).\hfill \end{array}$$

When $r=0$ and $2\ell +2\le J_k\le 3\ell +2$, that is, $k=i+1$ and i is even, by Chart 9, we have

$$\begin{array}{cc}\hfill \sum _{j=0}^{J_i-1}{m}_j^{\left(2\right)}& =\sum _{x=2\ell +2}^{J_k-1}{t}_{x,0}+\sum _{x=0}^{3\ell +2-J_k}{t}_{x,1}\hfill \\ \hfill & =\left(\frac{(J_k-1)J_k}{2}-\frac{(2\ell +1)(2\ell +2)}{2}\right)J_{i+2}\hfill \\ \hfill & +\frac{(3\ell +2-J_k)(3\ell +3-J_k)}{2}{J}_{i+2}+(3\ell +3-J_k)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}(5J_i-1)J_i{J}_{i+2}-4(3J_i-J_k)J_i{J}_{k-2}.\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}\hfill n_2(J_i,J_{i+2},J_{i+k})& =\frac{1}{2}(5J_i-1)J_{i+2}-4(3J_i-J_k)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}(5J_i{J}_{i+2}-J_i-J_{i+2}+1)-4(3J_i-J_k)J_{k-2}.\hfill \end{array}$$

When $r=0$ and $J_k\le 2\ell +1$, that is, $k=i$ or $k=i+1$ and i is odd, by Chart 10, we have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{J_i-1}{m}_j^{\left(2\right)}=(t_{J_k,0}+\cdots +t_{2\ell +1,0})+(t_{2\ell +2-J_k,1}+\cdots +t_{\ell ,1})\hfill \\ \hfill & =\left(\frac{(2\ell +1)(2\ell +2)}{2}-\frac{(J_k-1)J_k}{2}\right)J_{i+2}\hfill \\ \hfill & +\left(\frac{\ell (\ell +1)}{2}-\frac{(2\ell +1-J_k)(2\ell +2-J_k)}{2}\right)J_{i+2}+(J_k-\ell -1)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}(J_i+2J_k-1)J_i{J}_{i+2}-4(J_k-J_i)J_i{J}_{k-2}.\hfill \end{array}$$

Hence, we have

$$\begin{array}{cc}\hfill n_2(J_i,J_{i+2},J_{i+k})& =\frac{1}{2}(J_i+2J_k-1)J_{i+2}-4(J_k-J_i)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}\left((J_i+2J_k-1)J_{i+2}-J_i+1\right)-4(J_k-J_i)J_{k-2}.\hfill \end{array}$$

Theorem 8.

For $i,k\ge 3$, we have

$$\begin{array}{cc}\hfill & n_2(J_i,J_{i+2},J_{i+k})=\frac{1}{2}(5J_i{J}_{i+2}-J_i-J_{i+2}+1)(k\ge i+2),\hfill \\ \hfill & n_2(J_i,J_{i+2},J_{2i+1})\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{1}{2}\left(J_i+2J_{i+1}-1)J_{i+2}-J_i+1\right)-4(J_{i+1}-J_i)J_{i+1}\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ \frac{1}{2}(5J_i{J}_{i+2}-J_i-J_{i+2}+1)-4(3J_i-J_{i+1})J_{i-1}\hfill & \mathit{if}i\mathit{is}\mathit{even},\hfill \end{array}\right.\hfill \\ \hfill & n_2(J_i,J_{i+2},J_{2i})=\frac{1}{2}\left((3J_i-1)J_{i+2}-J_i+1\right),\hfill \\ \hfill & n_2(J_i,J_{i+2},J_{2i-1})\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{1}{2}\left((J_i+4J_{i-1}-1)J_{i+2}-J_i+1\right)-8J_i{J}_{i-3}\hfill & \mathit{if}i\mathit{is}\mathit{odd};\hfill \\ \frac{1}{2}\left(3J_i{J}_{i+2}-J_i-J_{i+2}+1\right)-4(5J_i-6J_{i-1})J_{i-3}\hfill & \mathit{if}i\mathit{is}\mathit{even}.\hfill \end{array}\right.\hfill \end{array}$$

When $r=⌊(J_i-1)/J_k⌋\ge 2$, that is, $k\le i-2$ ($i\ge 6$:even), we have

$$\begin{array}{cc}\hfill & n_2(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\frac{1}{2}\left((J_i+4J_k-1)J_{i+2}-J_i+1\right)-2\left(2r{J}_i-(r+3)(r-2)J_k\right)J_{k-2}.\hfill \end{array}$$

7. General p Case

We can continue the process with $p=3,4,\cdots$. When $r<p$, it is necessary to consider complicated case classification, but when $r\ge p$, the closed formula for p-genus can be obtained according to the above-mentioned certain rule. We have

$$\begin{array}{cc}\hfill & \sum _{j=0}^{J_i-1}{m}_j^{\left(p\right)}\hfill \\ \hfill & =\sum _{h=0}^{r-p-1}(t_{p{J}_k,h}+\cdots +t_{(p+1)J_k-1,h})\hfill \\ \hfill & +(t_{p{J}_k,r-p}+\cdots +t_{p{J}_k+\ell ,r-p})\hfill \\ \hfill & +(t_{(p-1)J_k+\ell +1,r-p+1}+\cdots +t_{p{J}_k-1,r-p+1})\hfill \\ \hfill & +(t_{(p-1)J_k,r-p+2}+\cdots +t_{(p-1)J_k+\ell ,r-p+2})\hfill \\ \hfill & +\cdots \hfill \\ \hfill & +(t_{\ell +1,r+p-1}+\cdots +t_{J_k-1,r+p-1})+(t_{0,r+p}+\cdots +t_{\ell ,r+p})\hfill \\ \hfill & =\frac{1}{2}\left((r-p)((2p+1)J_k-1)J_k+(J_i-(r-p)J_k-1)(J_i-(r-p)J_k)\right)J_{i+2}\hfill \\ \hfill & +\left(\frac{(r-p-1)(r-p)}{2}{J}_k+r(J_i-(r-p)J_k)\right)J_{i+k}\hfill \\ \hfill & =\frac{1}{2}(J_i+2p{J}_k-1)J_i{J}_{i+2}-2\left(2r{J}_i-(r+p+1)(r-p)J_k\right)J_i{J}_{k-2}.\hfill \end{array}$$

Hence, by (18), we have

$$\begin{array}{cc}\hfill & n_p(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\frac{1}{2}(J_i+2p{J}_k-1)J_{i+2}-2\left(2r{J}_i-(r+p+1)(r-p)J_k\right)J_{k-2}-\frac{J_i-1}{2}\hfill \\ \hfill & =\frac{1}{2}\left((J_i+2p{J}_k-1)J_{i+2}-J_i+1\right)-2\left(2r{J}_i-(r+p+1)(r-p)J_k\right)J_{k-2}.\hfill \end{array}$$

Theorem 9.

Let $i\ge 3$ and p be a non-negative integer. When $r=⌊(J_i-1)/J_k⌋\ge p$, we have

$$\begin{array}{cc}\hfill & n_p(J_i,J_{i+2},J_{i+k})\hfill \\ \hfill & =\frac{1}{2}\left((J_i+2p{J}_k-1)J_{i+2}-J_i+1\right)-2\left(2r{J}_i-(r+p+1)(r-p)J_k\right)J_{k-2}.\hfill \end{array}$$

(20)

Remark 6.

Concerning Fibonacci triples [16], for $r=⌊(F_i-1)/F_k⌋\ge p$ with $(r,p)\ne (0,0)$, we have

$$\begin{array}{cc}\hfill & n_p(F_i,F_{i+2},F_{i+k})\hfill \\ \hfill & =\frac{1}{2}\left((F_i+2p{F}_k-1)F_{i+2}-F_i+1\right)-\frac{1}{2}\left(2r{F}_i-(r+p+1)(r-p)F_k\right)F_{k-2}.\hfill \end{array}$$

(21)

8. Jacobsthal Polynomials

Jacobsthal numbers $J_n$ can be generalized as Jacobsthal polynomials $J_n\left(x\right)$, defined by the recurrence relation $J_n\left(x\right)=J_{n-1}\left(x\right)+x{J}_{n-2}\left(x\right)$ ($n\ge 2$) with $J_0\left(x\right)=0$ and $J_1\left(x\right)=1$ ([1], [Ch.44]). When $x=2$, $J_n=J_n\left(2\right)$ are the Jacobsthal numbers.

Let p be a given non-negative integer. It would be possible to give a complete explicit formula for a given $x=b$ and p, but it requires a very complicated case classification for each case, and it is more difficult to make a general statement. Nevertheless, it is possible to give a generalized formula only when $r\ge p$. The formula depends on the identity

$$J_{i+k}\left(b\right)=J_k\left(b\right)J_{i+2}\left(b\right)-b^2{J}_{k-2}\left(b\right)J_i\left(b\right).$$

Theorem 10.

Let $i,k\ge 3$ and p be a non-negative integer, and b be a positive integer. When $r=⌊(J_i\left(b\right)-1)/J_k\left(b\right)⌋\ge p$, we have

$$\begin{array}{cc}\hfill & g_p(J_i\left(b\right),J_{i+2}\left(b\right),J_{i+k}\left(b\right))\hfill \\ \hfill & =\left\{\begin{array}{c}(J_i\left(b\right)-r{J}_k\left(b\right)-1)J_{i+2}\left(b\right)+(r+p)J_{i+k}\left(b\right)-J_i\left(b\right)\hfill \\ \mathit{if}(r,p)=(0,0),\mathrm{or}(J_i\left(b\right)-r{J}_k\left(b\right))J_{i+2}\left(b\right)\ge b^2{J}_{k-2}\left(b\right)J_i\left(b\right);\hfill \\ (J_k\left(b\right)-1)J_{i+2}\left(b\right)+(r+p-1)J_{i+k}\left(b\right)-J_i\left(b\right)\hfill \\ \mathit{if}(J_i\left(b\right)-r{J}_k\left(b\right))J_{i+2}\left(b\right)<b^2{J}_{k-2}\left(b\right)J_i\left(b\right).\hfill \end{array}\right.\hfill \\ \hfill & n_p(J_i\left(b\right),J_{i+2}\left(b\right),J_{i+k}\left(b\right))\hfill \\ \hfill & =\frac{1}{2}\left((J_i\left(b\right)+2p{J}_k\left(b\right)-1)J_{i+2}\left(b\right)-J_i\left(b\right)+1\right)\hfill \\ \hfill & -\frac{b^2}{2}\left(2r{J}_i\left(b\right)-(r+p+1)(r-p)J_k\left(b\right)\right)J_{k-2}\left(b\right).\hfill \end{array}$$

Remark 7.

Since $F_n=J_n\left(1\right)$ are Fibonacci numbers, when $b=1$, Theorem 10 can be reduced to the Fibonacci results in [16].

It would be attractive to consider the numbers or polynomials with the same recurrence relations but different initial values. In ([1], [Ch.44]), Jacobsthal–Lucas polynomials $j_n\left(x\right)$ are introduced as $j_n\left(x\right)=j_{n-1}\left(x\right)+x{j}_{n-2}\left(x\right)$ ($n\ge 2$) with $j_0\left(x\right)=2$ and $j_1\left(x\right)=1$. $j_n=j_n\left(2\right)$ are called Jacobsthal–Lucas numbers and $L_n=j_n\left(1\right)$ are well-known Lucas numbers. However, explicit formula cannot be expressed only by the Jacobsthal–Lucas polynomials (or numbers), and the help of the Jacobsthal polynomials (or numbers) are also needed. The formula depends on the identity

$$j_{i+k}\left(b\right)=J_k\left(b\right)j_{i+2}\left(b\right)-b^2{J}_{k-2}\left(b\right)j_i\left(b\right).$$

Theorem 11.

Let $i,k\ge 3$ and p be a non-negative integer, and b be a positive integer. When $r=⌊(j_i\left(b\right)-1)/J_k\left(b\right)⌋\ge p$, we have

$$\begin{array}{cc}\hfill & g_p(j_i\left(b\right),j_{i+2}\left(b\right),j_{i+k}\left(b\right))\hfill \\ \hfill & =\left\{\begin{array}{c}(j_i\left(b\right)-r{J}_k\left(b\right)-1)j_{i+2}\left(b\right)+(r+p)j_{i+k}\left(b\right)-j_i\left(b\right)\hfill \\ \mathit{if}(r,p)=(0,0),\mathrm{or}(j_i\left(b\right)-r{J}_k\left(b\right))j_{i+2}\left(b\right)\ge b^2{J}_{k-2}\left(b\right)j_i\left(b\right);\hfill \\ (J_k\left(b\right)-1)j_{i+2}\left(b\right)+(r+p-1)j_{i+k}\left(b\right)-j_i\left(b\right)\hfill \\ \mathit{if}(j_i\left(b\right)-r{J}_k\left(b\right))j_{i+2}\left(b\right)<b^2{J}_{k-2}\left(b\right)j_i\left(b\right).\hfill \end{array}\right.\hfill \\ \hfill & n_p(j_i\left(b\right),j_{i+2}\left(b\right),j_{i+k}\left(b\right))\hfill \\ \hfill & =\frac{1}{2}\left((j_i\left(b\right)+2p{J}_k\left(b\right)-1)j_{i+2}\left(b\right)-j_i\left(b\right)+1\right)\hfill \\ \hfill & -\frac{b^2}{2}\left(2r{j}_i\left(b\right)-(r+p+1)(r-p)J_k\left(b\right)\right)J_{k-2}\left(b\right).\hfill \end{array}$$

9. Open Questions

It would be interesting and possible to find an explicit form of p-Frobenius number about the so-called Pell-type numbers, satisfying the recurrence relation $P_n^{*}=a{P}_{n-1}^{*}+P_{n-2}^{*}$ for an integer $a\ge 2$. One should not think the situation is similar. In this case, as the value of a increases, the value of $P_n^{*}$ increases even more rapidly, so it becomes more rapidly difficult to trace the actual situation. One of the difficulties is that there is no constant C that gives $P_{i+k}^{*}=P_k^{*}{P}_{i+2}^{*}-C{P}_{k-2}^{*}{P}_i^{*}$ for each fixed a, unlike in Jacobsthal.