2. Properties of Gilbreath Sequence
Definition 1. Let be a finite sequence of integers andwhere ; then, S is called a Gilbreath sequence iff . For example, let be a sequence of length and Gilbreath triangle of S:
…
Replacing these values gives
1
Let denote the set of all Gilbreath sequences of length n and be the set of all Gilbreath sequences. In the previous example, the first term of every sequence (except for the original sequence S) is equal to 1; then, .
Lemma 1. Let and be finite sequences of integers; then, .
Proof. Consider the Gilbreath triangle of S
…
where as a consequence of . Removing the last element of each sequence gives
…
which is a Gilbreath triangle of , as a consequence of ; then, . □
Definition 2. Let and be finite sequences of integers. Denote with the set of integers k such that .
The Gilbreath triangle of S is
…
where as a consequence of . The Gilbreath triangle of is
…
where as a consequence of . If , then .
Consider the equation
We will refer to Equation (
3) as a Gilbreath equation of
S. There are
values of
k that satisfy (
3); then,
is the set of all solutions for
k:
The largest value of
k that solves (
3) is
, and the smallest value is
. A remarkable relation is
Lemma 2. Let be a finite sequence of integers where ; then, .
Proof. Let , where . From Definition 2, if . Let now the sequence , from Definition 2, if . From the previous step, . Then, and . By induction, this can be proved for every element of S. If and the first element of S is an even number, then all the other numbers of the sequence will be odd. □
Lemma 3. Let be a finite sequence of integers where ; then .
Proof. This argument is the same argument as Lemma 2. □
Lemma 4. Let denote the sets and and let a finite sequence of integers where . Then, .
Proof. See Lemmas 2 and 3. □
Lemma 5. Let and be finite sequences of integers where . Then, Proof. See Definition 2 and Lemma 4, where the symbol ∧ is used to denote the truth-functional operator of logical conjunction AND [
4]. □
An important result regarding Equation (
4) follows from Lemma 5. Equation (
4) generates
solutions for a finite sequence
, where
. From Lemma 5, these solutions are only even or only odd if
is odd or even, respectively. Therefore, the number of distinct solutions generated by (
4) is
since solutions are divided between even and odd.
Theorem 1. Let and be finite sequences of integers; then, .
Proof. Suppose that
. Prove the right implication first. From Definition 2,
and from Lemma 5,
. Then,
. Prove the left implication by contradiction. Suppose that
but
. Then,
. The symbol ∨ is used to denote the truth-functional operator of logical disjunction OR [
4]. From Definition 2 and Lemma 5, it is not possible to have
if
. Then, it is true that
. □
Corollary 1. Let be a finite sequence of integers; then, .
Proof. As a consequence of Definition 2 and Equation (
4), each
m-th element of a sequence
S must be within the range between the upper and the lower sequences calculated on all the elements prior to the
m-th ones. From Definition 2 and according to the solution of Gilbreath Equation (
4), there cannot exist a Gilbreath sequence in which the
m-th element is larger than
, since
is the maximum value for the
m-th element. The same goes for
, since it is the smallest value for the
m-th element. □
3. Upper and Lower Bound Sequences
Let us now introduce the definition of two notable Gilbreath sequences. Let
be a finite sequence of integers; from (
4), any solutions of the Gilbreath equation cannot be greater than
, so the sequence
is the upper bound sequence for
S. Let the new sequence now be
and its upper bound sequence be
, and so on. Equally, let a finite sequence of integers be
; from (
4), any value of
k cannot be smaller than
and the new sequence
will have a lower limit for
, and so on. Then, it is possible to introduce the definition of the upper bound Gilbreath sequence and the lower bound Gilbreath sequence.
Definition 3. Let be a finite sequence of integers. Let us denote with the upper bound Gilbreath sequence for S and with the lower bound Gilbreath sequence for S:The following recursive definition holds:and It is also useful to define a notable sub sequence of and .
Definition 4. Let be a finite sequence of integers and its and . Let us define and as follows:The following recursive definition holds: From Theorem 1, and , while elements of and are all even or all odd; then, and .
From Definition 3, let
; then,
,
,
, and
.
Table 1 shows some examples of Gilbreath sequences and the closed form for
.
4. Gilbreath Polynomials
Definition 5. Let be the ordered sequence of the first m prime numbers and let be a function such that , where ; then, is called m-th Gilbreath polynomial.
Through simple algebra, one can prove that for the ordered sequence of the first
m prime numbers,
are represented in
Table 2. This provides important information about sequence A347924 [
5], which is the triangle read by rows, where row
m is the
m-th Gilbreath polynomial and column
n is the numerator of the coefficient of the
k-th degree term. According to
Table 2, this sequence contains the integer term of every
m-th Gilbreath polynomial. The related sequence A347925 [
6] contains the lowest common denominator of
m-th Gilbreath polynomial. It is the sequence of denominators of the polynomials in
Table 2.
Relation between Gilbreath Polynomials and
Gilbreath polynomials are closely related to prime numbers and . Let a finite sequence of integers , Theorem 1 and the following. The relationship must be true; otherwise, it would not be true that . As a consequence of Lemma 5, for all elements subsequent to , the absolute difference of two successive elements must be an integer multiple of 2 so as to maintain the absolute difference of two successive elements as an even value. So, if the first element in the sequence is even, the subsequent elements must be odd, and if the first element is odd, the subsequent elements must be even.
Let
be a Gilbreath sequence formed by the first two prime numbers. From (
5),
and from Theorem 1,
. By definition of
P,
. Since
, it is certainly true that
. The left inequality is proved for
and it is easy to prove for every
n. The proof of
is trivial and holds for all prime numbers, hence
. Given Gilbreath polynomials in Definition 5,
, then
The left side of (
6)
consists of a Gilbreath polynomial conjecture whose solution implies
. Unfortunately, bounds for
are not enough good to prove (
7); however, this opens the way for a new approach to the
[
7,
8,
9,
10].