Gilbreath’s sequences and proof of conditions for Gilbreath’s conjecture

The conjecture attributed to Norman L. Gilbreath, but formulated by Francois Proth in the second half of the 1800s, concerns an interest-ing property of the ordered sequence of prime numbers P . Gilbreath’s conjecture stated that, computing the absolute value of diﬀerences of consecutive primes on ordered sequence of prime numbers, and if this calculation is done for the terms in the new sequence and so on, every sequence will starts with 1. In this paper is deﬁned the concept of Gilbreath’s sequence, Gilbreath’s triangle and Gilbreath’s equation. On the basis of the results obtained from the proof of properties, an inductive proof is produced thanks to which it is possible to establish the necessary condition to state that the Gilbreath’s conjecture is true.


Introduction to Gilbreath's conjecture
Let the ordered sequence P = {2, 3, 5, 7, 11, 13, 17, ...} = {p 1 , ..., p n , ...} formed by prime numbers, and set p b a = |p b−1 a+1 − p b−1 a | where b 1. Gilbreath conjectured that every term p b 1 = 1. In this notation, the elements of P should be indicated with {p 0 1 , ..., p 0 n , ...}. For brevity, the superscript with b = 0 is omitted. It is likely that this conjecture is satisfied by many other sequences of integers, so it is necessary to define the general properties of all sequences that satisfy this conjecture. The first term of every sequence is equal to 1, hence S ∈ G 7 . Theorem 2. Let a sequence S = {s 1 , ..., s n } ∈ G n and a sequence S = {s 1 , ..., s n , k}, than S ∈ G n+1 ⇔ k ∈ K S .
Proof. The Gilbreath's triangle associated with S:  Proof. The Gilbreath's equation is a 2 n degree equation, then there are 2 n value of k that satisfy the equation (1). The solutions are: With respect to K S , the largest value that solves the equation is max K S = s n−1 1 +s n−2 2 +s n−3 3 +s n−4 4 +...+s 1 n−1 +s n +1 and the smallest value that solves the equation is min Corollary 2. Let a sequence S = {s 1 , ..., s n } ∈ G n and a sequence S = {s 1 , ..., s n , s n }, than S ∈ G n+1 .
Proof. The Gilbreath's equations of S with k = s n is: It is useful, for the following proofs to introduce the definition of two important Gilbreath's sequences. Let a sequence S = {s 1 , ..., s n }, from relation (2), any value of k cannot be greater than max K S , so the sequence {s 1 , ..., s n , max K S } is the upper bound sequence for the sequence {s 1 , ..., s n }.
The new sequence S = {s 1 , ..., s n , max K S , k} will have the upper limit for k = max K {s 1 ,...,sn,max K S } . Equally, let a sequence S = {s 1 , ..., s n }, from relation (2), any value of k cannot be smaller than min K S and the new sequence S = {s 1 , ..., s n , min K S , k} will have the lower limit for k = min K {s 1 ,...,sn,min K S } . From this example, it is now possible to introduce the definition of upper bound sequence and lower bound sequence.
From the previous step, s 1 ±1 ∈ 2Z+1, hence 1+s 1 ±1 ∈ 2Z and 1+s 1 ±1±1 ∈ 2Z+1. Iteratively, this can be proved for every element of S. Hence if S ∈ G n and the first element of S is an even number, then all the other numbers of the sequence must be odd. Proof. Let S 1 = {s 1 }, where s 1 ∈ 2Z + 1. From theorem 2, corollary 1: From the previous step, s 1 ±1 ∈ 2Z, hence 1+s 1 ±1 ∈ 2Z+1 and 1+s 1 ±1±1 ∈ 2Z. Iteratively, this can be proved for every element of S. Hence if S ∈ G n and the first element of S is an odd number, then all the other numbers of the sequence must be even.  Proof. From theorem 2, corollary 1, there are 2 n values of k that satisfy S ∈ G n+1 and from definition 2 and definition 3, min K S k max K S . K S is defined as the set of all solutions of k, hence it contais elements between min K S and max K S . From lemma 3 it has already been shown that if s 1 ∈ 2Z, than s a ∈ 2Z+1, ∀a > 1 and if s 1 ∈ 2Z+1, than s a ∈ 2Z, ∀a > 1, the theorem is proved from theorem 2, definition 2, definition 3 and lemma 3.
From lemma 4 is proved an important result regarding (2). (2) generates 2 n solutions for a sequence S = {s 1 , ..., s n , k}, but from lemma 43 it has been proved that these solutions are only even or only odd according to the nature of the sequence. Therefore, the number of distinct solutions generated by (2) is 2 n−1 since half solutions between min K S and max K S are equally divided between even and odd: dim K S = 2 n−1 .
Theorem 3. Let a sequence S = {s 1 , ..., s n } ∈ G n and S = {s 1 , ..., s n , k}, Proof. From definition 2 and definition 3, k ∈] min K S ; max K S [ and from lemma 4, k ∈ 2Z No remarkable expression was found to be to analytically define the trend of U S and L S for a generic sequence S but it was observed that the exponential trend is preserved. However, this trend varies with the number of terms of U S and L S so it does not seem possible to establish what will be the n + 1-th term of U S and L S given the previus n terms through an analytical formula. However, it is always possible use the recursive espression (2).
If it is true that exponential trend is preserved, elements of U S can be written in the form u Sn = αe βn or log u Sn = log α + βn.
The best fit for a dataset D = {d 1 , ..., d n } in a linear regression model is and the coefficient of determination is Note that in D if |d a | < |d a+1 | and d a < 0, it is not possible to calculate log d b , where b > a. To avoid this problem the transformation After that, the fitting curve will be d n = αe βn − d 1  As explained above, the addition of a term to U S leads to new values of α and β, therefore this analysis can be carried out without pretending to evaluate the n + 1-th element of a given U S of length n.
The numerical analysis of the values of the upper limit sequence was added only to show that no analytical formula has been found for the generations of the values of this sequence, with exception of (2).
3 Proof of conditions for P = {p1, ..., p n } Let S = {s 1 , ..., s n } = {f (n)}, from theorem 1 and 2, is true the following. The relationship s 2 = s 1 ± 1 must be true, otherwise it would not be true that s 1 1 = 1, hence f (2) = f (1) ± 1. As a consequence of theorem 3, for all elements subsequent to s 1 , 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.
As a consequence of theorem 2, solution of the Gilbreath's equation (2) and definition 2 and definition 3: each n-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 n-th ones. Hence, from theorem 2 and according to the solution of the Gilbreath's equation at (2), cannot exists a Gilbreath's sequence in which the n-th is larger than max K {s 1 ,...,s n−1 } , since max K {s 1 ,...,s n−1 } is the maximum value that the n-th value can take according to (2). The same goes for min K {s 1 ,...,s n−1 } , since it is the smallest value that the n-th value can take, according to (2). Hence: l {s 1 ,...,s n−1 } n f (n) u {s 1 ,...,s n−1 } n (14) Following the results obtained in the previous paragraphs about Gilbreath's sequence and Gilbreath's equation, let proceed discussing the Gilbreath's conjecture. The results obtained so far will be used to establish if theorem (4) is true for an ordered sequence of prime numbers P .