2. Properties of Gilbreath sequence
Definition 1.
Let be a finite sequence of integers and
where , then S is called Gilbreath sequence iff .
For example, let be a sequence of length and Gilbreath triangle of S
…
Replacing values gives
1
Let denote the set of all Gilbreath sequences of length n and 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 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 .
Gilbreath triangle of S is
…
where as a consequence of . Gilbreath triangle of is
…
where as a consequence of . If , then .
We will refer to equation (
3) as 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. Same argument as lemma 3. □
Lemma 4. Let denote the sets and and let a finite sequence of integers where . Then .
Proof. Lemma 2 and lemma 3. □
Lemma 5. Let and be finite sequences of integers where . Then
.
Proof. Definition 2 and lemma 4. □
An important result regarding equation (
4) follows from lemma 5. (
4) generates
solutions for a finite sequence
where
. From lemma 5, these solutions are only even or only odd if
is odd of 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 . From definition 2 and lemma 5 it is not possible to have if . Then it is true . □
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 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. □
4. Gilbreath polynomials
Definition 5. Let be the ordered sequence of 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 first
m prime numbers,
are represented in
Table 2. This provides important information about sequence A347924 [
7] 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 polynomials. 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.
4.1. Relation between Gilbreath polynomials and
Gilbreath polynomials are closely related to prime numbers and . Let a finite sequence of integers , from theorem 1 is true 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
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
Left side of (
6)
consists of 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
[
1,
2,
4,
5].