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. □