1. Introduction
We consider the sequence
, satisfying
where
u and
v are positive integers with
. The values of
depend on the values of
u and
v. If
,
is the
n-th Fibonacci number [
1]. If
and
,
is the
n-th Jacobsthal number [
2,
3]. If
and
,
is the
n-th Pell number [
4]. However, for simplicity, if we do not specify the values of
u or
v, we will simply write
for
.
This type of number sequence has been well known to many people by Horadam’s series of studies ([
5,
6,
7,
8,
9]) in the 1960s. Because of this fact, this sequence is sometimes called the
Horadam sequence. Horadam himself used the recurrence relation
. But recently more people (see, e.g., [
10,
11]) have used the recurrence relation
and such works are still due to Horadam. In general, the initial values are arbitrary, but because of some simplifications, we set
and
. According to [
6], this sequence has long exercised interest, as seen in, for instance, Bessel-Hagen [
12], Lucas [
13], and Tagiuri [
14], and, for historical details, Dickson [
15]. However, it is deplorable that quite a few papers are publishing results that have already been obtained by these authors as new results, either because they are unaware of their or the following important results, or even if they are ignoring them.
Given the set of positive integers
(
), for a nonnegative integer
p, let
be the set of integers whose nonnegative integral linear combinations of given positive integers
are expressed in more than
p ways. For a set of nonnegative integers
, the set
is finite if and only if
. Then there exists the largest integer
in
, which is called the
p-Frobenius number. The cardinality of
is called the
p-genus and is denoted by
. The sum of the elements in
is called the
p-Sylvester sum and is denoted by
. This kind of concept is a generalization of the famous Diophantine problem of Frobenius since
is the case when the original Frobenius number
, the genus
and the Sylvester sum
are recovered. We can call
the
p-numerical semigroup. Strictly speaking, when
,
does not include 0 since the integer 0 has only one representation, so it satisfies simply additivity, and the set
becomes a numerical semigroup. For numerical semigroups, we refer to [
16,
17,
18]. For the
p-numerical semigroup, we refer to [
19].
We are interested in finding any closed or explicit form of the
p-Frobenius number, which is even more difficult when
. For three or more variables, no concrete example had been found. Most recently, we have finally succeeded in giving the
p-Frobenius number as closed-form expressions for the triangular number triplet ([
20]), for repunits ([
21,
22]).
In this paper, we study the
p-numerical semigroups of the triple
for integers
. We give explicit closed formulas of the
p-Frobenius numbers and
p-genus of this triple. Note that the special cases for Fibonacci [
1], Pell [
4], and Jacobsthal triples [
2,
3] have already been studied.
2. Preliminaries
We introduce the Apéry set (see [
23]) below in order to obtain the formulas for
,
, and
technically. Without loss of generality, we assume that
.
Definition 1.
Let p be a nonnegative integer. For a set of positive integers with and we denote by
the p-Apéry set of A, where each positive integer satisfies the conditions:
Note that is defined to be 0.
It follows that for each
p,
Even though it is hard to find any explicit form of
as well as
and
, by using convenient formulas established in [
24,
25], we can obtain such values for some special sequences
after finding any regular structure of
. One convenient formula is on the power sum
by using Bernoulli numbers
defined by the generating function
and another convenient formula is on the weighted power sum ([
26,
27])
by using Eulerian numbers
appearing in the generating function
with
and
. Here,
is a nonnegative integer and
. From these convenient formulas, many useful expressions are yielded as special cases. Some useful ones are given as follows. The formulas (3) and (4) are entailed from
and
, respectively.
Lemma 1.
Let κ, p and μ be integers with and . Assume that . We have
Remark 1.
When , the formulas (2), (3) and (4) reduce to the formulas by Brauer and Shockley [28] [Lemma 3], Selmer [29] [Theorem], and Tripathi [30] [Lemma 1]1, respectively:
where () with .
3. Main Results
We use the following properties repeatedly. The proof is trivial and omitted.
First of all, if
i is odd and
then by (
1) and (7),
Hence,
. Therefore, from now on, we consider the case only when
i is even and
k is odd, or when
i is odd, with
.
3.1. The Case k Is Odd
When
k is odd, we choose non-negative integers
and
as
where
if
due to (
5), otherwise
is the largest integer, satisfying
More directly, when
i is even (and
k is odd),
When
i is odd (and
k is odd),
Note that if
([
2]), then always
.
In particular, if
i is even and
If
, then by (
5)
. So, when
i is even, by (8)
(8). Thus, we get
When
and
i is odd, by
and
, there exists an integer
h such that
. By
,
. Thus,
Thus, we get
We use the following identity.
Proof. By (
1) and (7) together with (
9), we get
□
Assume that
(The case
is discussed later). Then the elements of the (0-)Apéry set are given in
Table 1. Here, we consider the expression
or simply the position
.
We shall show that all the elements in
Table 1 constitute the sequence
in the vertical
y direction. However, if
i is odd and
i is even, the situation of this sequence is different. In short, if
i is odd, the sequence appears continuously, but if
i is even, the sequence is divided into
u subsequences.
First, let
i be odd. Then by
, we have
By (7), we get
Hence,
Thus, the element at
(
) cannot be an element of
but
as the same residue modulo
, where
. Next, by Lemma 3, we have
Thus, the element at
(
) cannot be an element of
but
.
Therefore, the sequence
is divided into the longer parts with length
and the shorter parts with length
: Namely, the longer part is of the subsequence
with the next element at
. The shorter part is of the subsequence
with the next element at
. Since
, all elements in
are different modulo
.
Next, let
i be even. Then by
, we have
Hence,
with
(
). By the determination of
in (
11), we see that
. So, by using the relations (
14). Thus, each subsequence is given as the following points. For
with next element is at
, coming back to the first one, because of Lemma 3. In addition, by (8), all terms of the above subsequence are
Since
, this is equivalent to
(
). Therefore, there is no overlapped element among all subsequences. By (
9), the total number of terms in each subsequence is
as expected.
By
Table 1, the candidates of the largest element of
are at
or at
. Since
is equivalent to
, by Lemma 1 (
2), if
, then
If
, then
3.1.1. The Case k Is Odd with
When
k is odd and
, we get
and
. Hence, the elements of the (0-)Apéry set are given in
Table 2.
Similarly to the case , when i is odd, so , the sequence simply becomes one sequence by combining all the subsequences with length and with length . When i is even, so , the sequence consists of u subsequences with the same length .
By
Table 2, the largest element of
is at
. Hence,
In fact, this is included in the case where
and
.
3.2. The Case k Is Even
When
k is even (so
i is odd), we choose non-negative integers
q and
r as
where
. Note that
is an integer for even
k.
Note that
because otherwise
i is also even. Then the elements of the (0-)Apéry set are given in
Table 3.
Similarly to the case where
k is odd in (
14), we have
Thus, the element at
(
) cannot be an element of
but
as the same residue modulo
. The sequence
is divided into the longer parts with length
and one shorter part with length
r: Namely, the longer part is of the subsequence
with the next element at
. One shorter part is of the subsequence
with the next element at
. Notice that similarly to Lemma 3, we have
Since
, all elements in
are different modulo
. Then by
, we have
By
Table 3, the candidates of the largest element of
are at
or at
. Since
is equivalent to
, by Lemma 1 (
2), if
, then
If
, then
Notice that may occur in some cases. For example, . In this case, both of the two formulas are valid, yielding the Frobenius number .
4. The Case where
It is important to see that the elements of are determined from those of .
4.1. When k Is Odd
4.1.1. When
The corresponding relations from
to
are as follows. See
Table 4.
[The first
u rows]
by Lemma 3 and
respectively. Note that when
, the second corresponding relation does not exist. This also implies that all the elements at
and
can be expressed in terms of
in at least two ways.
[Others]
by the identity (
13). This also implies that all the elements at
can be expressed in at least two ways.
By
Table 4, there are four candidates to take the largest value of
. Namely, the values at
If
, one of the elements at
and at
is the largest. In this case, if
, then
If
, then
If
, one of the elements at
and at
is the largest. In this case, if
, then
If
, then
Examples.
When
, the first identity is applied:
Indeed, there are two representations in terms of
as
which is the largest element of
. In fact, the second, the third and the fourth identities yield the smaller values
respectively.
When
, the second identity is applied:
In fact, the first, the third and the fourth identities yield the smaller values
respectively.
When
, the third identity is applied:
In fact, the first, the second and the fourth identities yield the smaller values
respectively.
When
, the fourth identity is applied:
In fact, the first, the second and the third identities yield the smaller values
respectively.
4.1.2. When
The similar corresponding relations to the case
are also applied for
. When
, the elements of the first
u rows of the main area (the second block from the left) correspond to fill the gap below the left-most block:
The other elements of the main area correspond to those in the block immediately to the right to go up the
u row:
The elements of the stair areas correspond to those in the block immediately to the right in the form as it is to go up the
row:
See
Table 5. We can also show that all these elements have at least three distinct representations in terms of
.
From
Table 5, there are six candidates to take the largest element of
. These elements are indicated as follows.
In conclusion, if
and
, then
If
and
, then
If
and
, then
If
and
, then
In general, for an integer
, it is sufficient to compare two elements at both ends. See
Table 6. If
and
, then
If
and
, then
If
and
, then
If
and
, then
The positions of the elements of
below the left-most block and the positions of
in the right-most block are arranged as shown in
Table 6.
This situation is continued as long as . However, when , the shape of the block on the right side collapses. Thus, the regularity of taking the maximum value of is broken. Hence, the fourth case holds until and other cases hold for .
Theorem 1.
Let i be an integer and k be odd with . Let and be determined as (9) and (10). For , if and , then
If and , then
If and , then
If and , then for
Examples.
When
, the first identity is applied. Since
and
, for
we have
Namely, the corresponding element for each integer is at
(
). However, for
, the
p-Frobenius numbers can be computed neither by the above formula nor by any other closed formulas. Namely, the real value is
, corresponding to
, though the formula gives 1274931, corresponding to
.
4.2. When k Is Even
4.2.1. When
Similarly to the odd case where k is odd, the elements of can be determined from those of . When , there are corresponding relations as follows.
[The first row
]
with
due to (
15). Note that when
the second corresponding relation does not exist. This also implies that all the elements at
and
can be expressed in terms of
in at least two ways.
[Others]
by the identity (
13). This also implies that all the elements at
can be expressed in at least two ways.
By
Table 7, there are four candidates to take the largest value of
. Namely, the values at
If
, one of the elements at
and at
is the largest. In this case, if
, then
If
, then
If
, one of the elements at
and at
is the largest. In this case, if
, then
If
, then
4.2.2. When
The situation is similar for
. From
Table 8, there are six candidates to take the largest element of
. These elements are indicated as follows.
In conclusion, if
and
, then
If
and
, then
If
and
, then
If
and
, then
In general, for an integer
, it is sufficient to compare two elements at both ends. See
Table 9. If
and
, then
If
and
, then
If
and
, then
If
and
, then
The positions of the elements of
below the left-most block and the positions of
in the right-most block are arranged as shown in
Table 6.
This situation is continued as long as . However, when , the shape of the block on the right side collapses. Namely, we cannot take the value at . Thus, the regularity of taking the maximum value of is broken. Hence, the fourth case holds until , and other cases hold for .
Theorem 2.
Let i be an integer and k be even with . Let q and r be determined as (15). For , if and , then
If and , then
If and , then
If and , then for
Example
When
, we have
and
. So, the elements of
, where
, are given as in
Table 10. The largest element is at
, which comes from the second identity. Thus,
Notice that the right-most element is at and the block of the right side is empty. Therefore, the formula does not hold for . In fact, , corresponding to , though the formula gives 63914, corresponding to .
5. p-Genus
5.1. The Case Where k Is Odd
Let
k be odd. For a non-negative integer
p, the areas of the
p-Apéry set can be divided into three parts: the stairs part (left), the stairs part (right) and the main part. By referring to
Table 6 (with
Table 4 and
Table 5), we can compute
Here, we used the relation (
9) to simplify the expression. In addition, by
, we have
By Lemma 1 (3), we have
Since the
z value of the right-most side must be non-negative,
. Namely, the above formula is valid for
.
Example
When
, by
for
we have for
However, for
, the
p-genus cannot be obtained by the above formula. The real values are given by
though the formula gives
5.2. The Case Where k Is Even
Similarly to the case for
k is odd, when
k is even, by referring to
Table 9 (with
Table 7 and
Table 8), we can compute
Here, we used the relation (
15) to simplify the expression. In addition,
are all positive integers. By Lemma 1 (3), we have
Theorem 3.
Let i and k be integers with and . When k is odd, for we have
where and are given in (9). When k is even (and i is odd), for we have
where q and r are given in (15).
Example
Let
. So,
. Then for
by the formula we have
However, contrary to the fact that
, the formula gives 50746.
6. Final Comments
The original numbers studied by Horadam satisfy the recurrence relation
. From this point of view, almost all the above identities hold by replacing
v by
, though the condition
is necessary. For example, the identities of (7) and (8) are replaced by
respectively. For example, when
, by
for
by the first identity of Theorem 1, we have
When
, by
for
by the first identity of Theorem 2, we have
Horadam also studied the number with arbitrary initial values and . However, with arbitrary initial values many identities (e.g., (7)) do not hold as they are. Hence, the situation becomes too complicated.
Author Contributions
writing—original draft preparation, T.K.;writing—review and editing, 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 author declares no conflict of interest.
Notes
1 |
There was a typo, but it was corrected in [ 31]. |
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Table 1.
for odd k
Table 1.
for odd k
Table 2.
when
Table 2.
when
Table 3.
for even k
Table 3.
for even k
Table 4.
() for odd k
Table 4.
() for odd k
Table 5.
() for odd k
Table 5.
() for odd k
Table 6.
for odd k
Table 6.
for odd k
Table 7.
() for even k
Table 7.
() for even k
Table 8.
() for even k
Table 8.
() for even k
Table 9.
for even k
Table 9.
for even k
Table 10.
for
Table 10.
for
|
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