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Background

To create a algebraic formulation of Gilbreath's conjecture we will do the following. Using the notation $i=1$ and $p_k$ is the $k$'th prime:

$$ x^{p_i} + x^{p_{i+1}} = x^{p_i} (1+ x^{|p_i - p_{i+1}|})$$

$$ \implies \frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} = (1+ x^{|p_i - p_{i+1}|})$$

However for $i+1$ we have:

$$ \implies \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} = (1+ x^{|p_{i+1} - p_{i+2}|})$$

Adding the above two equations we have:

$$ \implies \frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2 = x^{|p_{i+1} - p_{i+2}|} + x^{|p_i - p_{i+1}|}$$

Now assuming Gilbreath's conjecture to be true and $ |a-b|=1 \implies a = b + O(1)$

$$ \implies \frac{\frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2}{x^{|p_{i} - p_{i+1}| +O(1)}} = 1+ x^{||p_{i+1} - p_{i+2}|-|p_i - p_{i+1}||}$$

However for $i+1$ we have:

$$ \implies \frac{\frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} + \frac{x^{p_{i+2}} + x^{p_{i+3}}}{x^{p_{i+2}}} -2}{x^{|p_{i+1} - p_{i+2}| +O(2)}} = 1+ x^{||p_{i+2} - p_{i+3}|-|p_{i+1} - p_{i+2}||} *$$

Adding the above two equations we have:

$$ \implies \frac{\frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2}{x^{|p_{i} - p_{i+1}| +O(1)}} + \frac{\frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} + \frac{x^{p_{i+2}} + x^{p_{i+3}}}{x^{p_{i+2}}} -2}{x^{|p_{i+1} - p_{i+2}| +O(2)}} - 2 = x^{||p_{i+1} - p_{i+2}|-|p_i - p_{i+1}||}+ x^{||p_{i+2} - p_{i+3}|-|p_{i+1} - p_{i+2}||}$$

$$ \implies \frac{\frac{\frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2}{x^{|p_{i} - p_{i+1}| +O(1)}} + \frac{\frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} + \frac{x^{p_{i+2}} + x^{p_{i+3}}}{x^{p_{i+2}}} -2}{x^{|p_{i+1} - p_{i+2}| +O(2)}} - 2}{x^{||p_{i+1} - p_{i+2}|-|p_i - p_{i+1}||+O(1)}}= 1+x^{|\dots|} = 1+x$$

*Why? Because using Gilberts conjecture:

$$ ||p_{i+1} - p_{i+2}|-|p_i - p_{i+1}|| = ||p_{i+2} - p_{i+3}|-|p_{i+1} - p_{i+2}|| + O(1)$$

But,

$$ \implies 1 + O(1) = ||p_{i+2} - p_{i+3}|-|p_{i+1} - p_{i+2}|| $$

Conclusion

We notice the following pattern emerging:

Generation 1:

$$ \frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} = (1+ x) $$

Generation 2:

$$ \frac{\frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2}{x^{|p_{i} - p_{i+1}| +O(1)}} = 1+ x$$

Generation 3:

$$ \implies \frac{\frac{\frac{x^{p_i} + x^{p_{i+1}}}{x^{p_i}} + \frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} -2}{x^{|p_{i} - p_{i+1}| +O(1)}} + \frac{\frac{x^{p_{i+1}} + x^{p_{i+2}}}{x^{p_{i+1}}} + \frac{x^{p_{i+2}} + x^{p_{i+3}}}{x^{p_{i+2}}} -2}{x^{|p_{i+1} - p_{i+2}| +O(2)}} - 2}{x^{||p_{i+1} - p_{i+2}|-|p_i - p_{i+1}||+O(1)}}= 1+x$$

Generation 4:

$$\vdots$$

Question

Does this already exist? Can any sort of distribution be obtained by this formulation of Gilbreath's conjecture? How does that distribution compare with Prime Number Theorem?

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    Reason for downvote?2017-02-16
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    Sorry but this question is still very fuzzy and unclear. It took me a couple readings to see that you really do mean to fix $i=1$. Why introduce the variable at all? Why the strange use of $O(n)$ to apparently mean "the sum of $n$ values $\pm1$? With each generation you generate exponentially more of these indefinite terms, so what hope is there of extracting distributional information after introducing more uncertainty than one already has from Gilbreath's conjecture itself?2017-02-23
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    Note that Gilbreath's conjecture by itself conveys very little information as a great many integer sequences satisfy the conjecture. Even the sequence $2^n+1$ satisfies it, even though it's *extremely* far away from the density of primes.2017-02-23

2 Answers 2

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This is rather a comment than an answer.

I just looked at it using Pari/GP having the first 1 000 000 primes at hand. Because a change of the leading $a_{r,1}$ to something else needs that the neighboured value $a_{r,2}$ is greater than 2, so must be some element from $\{4,6,8,10,...\}$ (here the index $r$ should indicate the iteration-number). This is the same rationale which A. Odlyzko has already applied in his article.

Since the iterations have the tendency to flatten the (absolute) differences downwards to $0$ and $2$ (which is, as I understand, you want to show) I looked at the index, where the first time $k=1,2,3,... $ a value greater than 2 occurs in some $a_{r,k}$. This gives the following picture:
picture

A rescaled version gives a nearly linear relation for the asymptotic:
picture

Doing a regression at that transformations of the iteration-index and of the position $y_{r}$ of the first occurence of $a_{r,k}>2$ I got the rough estimate $$ y_r \approx \large{ b ^ {(r^ { \;c })} }$$ where $b\approx 4.38972227354$ and $ c \approx 0.444088715260$

If such a relation can be established this would prove the Gilbreath's conjecture...

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There is practically no useful information about the sequence $p_n$ that can be deduced from the statement of Gilbreath's conjecture. Note that the sequence of powers of $2$ also has the same column of $1$s in its table of absolute differences. So does the sequence consisting of $2$ followed by all odd integers greater than $2$.

Any property that holds as a result of this arcane analysis must also hold for sequences as dense as $2n$ and as sparse as $2^n$. It would therefore have essentially no overlap with the Prime Number Theorem, which is purely about the density of primes.