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Background

Similar to Gilbreath's conjecture I tried to create my own conjecture.

We start with the primes:

$$ 2,3,5,7,11,13 $$

We then take the absolute difference between the terms side by side and call this the "same row difference step": $$ 2,3,5,7,11,13,\dots $$

$$ 1,2,2,4,2,\dots$$

Now this is where I differ from the standard conjecture. We do the absolute difference in terms (take the absolute difference of the first term of the first row with the first term of the second row. Repeat for the $n$'th term.) of the row and the row above it and call this the "double row difference step".

$$ 2,3,5,7,11,13,\dots $$

$$ 1,2,2,4,2,\dots$$

$$ 1,1,3,3,9,\dots$$ Again now we repeat the "same row difference step":

$$ 2,3,5,7,11,13,\dots $$

$$ 1,2,2,4,2,\dots$$

$$ 1,1,3,3,9,\dots$$

$$ 0,2,0,6,\dots$$

Now we again do the "double row difference step":

$$ 2,3,5,7,11,13,\dots $$

$$ 1,2,2,4,2,\dots$$

$$ 1,1,3,3,9,\dots$$

$$ 0,2,0,6,\dots$$

$$ 1,1,3,3,\dots$$

After alternating between the" double row difference step" and the "same row difference step" we notice a alternating $0,1$ sequence.

Conjecture

The first term is always alternating between $0$ and $1$ from the $3$rd row onwards.

Algebraic Representation

We will now represent the above steps algebraically:

Let $x^\lambda + y^\lambda = 2 I$

Where $\lambda$ is any arbitrary integer and $I $ is the identity matrix. $x= I - \epsilon$ and $y= I + \epsilon$ where $\epsilon$ is a nil potent matrix $\epsilon^2 = 0$. Now consider the series:

$$ K = sx^2+s^2x^3 + s^3x^5 + \dots + s^ix^{p_i} $$

Where $p_i$ is the $i$'th prime.

  1. Same row difference

(Representing the "same row difference step")

Multiplying $s$ both sides and Substracting:

$$ sK =0+s^2 x^2+s^3x^3 + \dots + s^{i+1}x^{p_i} $$ $$ K = sx^2+s^2x^3 + s^3x^5 + \dots + s^ix^{p_i} $$

$$ K(s-1)+sx^2= s^2x^2(1-x) +s^3x^3(1-x^2) + \dots$$

Using the relation between $x$ and $y$: $$\implies 2Ks -K +sx^2 = s^2x^2y +s^3x^3y^2 + \dots$$

  1. Double Row Difference

(Representing the "double row difference step")

Using $xy=I$:

$$\implies 2Ks -K +sx^2 = s^2x +s^3x + \dots$$ Again multiplying with $s$ and subtracting:

$$ 2Ks^2 -Ks +s^2x^2 = 0+ s^3x +s^4x + \dots$$ $$ 2Ks -K +sx^2 = s^2x +s^3x + \dots$$

$$\vdots$$

Questions

If the conjecture is true what does it mean in terms of the $K$ series? Is the conjecture correct? Can someone computationally check it?

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  • 1
    $\begingroup$ Reason for downvote? $\endgroup$ – drewdles Dec 19 '17 at 19:12
  • $\begingroup$ i do need more explanation of how you got 1,2,2,4,4. However, I am not familiar with Goldbarch's Conjecture so I might just be missing knowledge. Adding in the actual arithmetic for that step in the example might be helpful. As for this question being clear, it does sound clear. I understand the basic thing your asking. I just don't have enough knowledge to say much. $\endgroup$ – The Great Duck Dec 19 '17 at 21:08
  • $\begingroup$ You ought to have $13-11=2$, not $4$. $\endgroup$ – Simply Beautiful Art Dec 19 '17 at 21:10
  • $\begingroup$ @SimplyBeautifulArt you are correct but the conjecture still holds! $\endgroup$ – drewdles Dec 19 '17 at 21:14
  • $\begingroup$ @Typhon it should be$1,2,2,4,2 $ we get it by taking the absolute difference of the neighbors of the above row: $1,2,2,4,2 = |3-2|, |5-3|, |7-5|, |11-7|, |13-11|$ $\endgroup$ – drewdles Dec 19 '17 at 21:18
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As posted by a kind stranger on the internet the conjecture fails at row $104$.

Data

But why does it happen for the first $103$ lines is still a mystery to me ...

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    $\begingroup$ Just noted it malfunctions due to twin primes $239$ and $241$ at row $104$ ... $\endgroup$ – drewdles Dec 19 '17 at 23:24

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