Identity from Gilbreath's conjecture?

Question

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?

Answer 1

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.

Answer 2

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:

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

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...