1
$\begingroup$

Background and Question

We all know the Gilbreath conjecture. Inspired by that I wondered if I could do my own series analysis. If one proceeds to take absolute difference between row elements and then proceeds to subtract it a (hopefully) non-random pattern will emerge. My question is can someone give an (asymptotic) estimate of the growth of the left most terms above the dotted lines when one considers the primes?

Example:

Consider the primes:

$$ 2,3,5,7,11,13,\dots$$ $$---------$$ $$ 1,2,2,4,2,\dots$$ $$ 1,1,3,3,9\dots$$ $$------$$ $$0,2,0,6$$ $$1,-1,3,-3$$ $$-----$$ $$2,4,6$$ $$1,-5,-3$$ $$---$$

Why is this question interesting to me? (Extra)

Even though these operations seem arbitrary they aren't as they can be represented as shown below:

Let $x= 1 - \epsilon$ and $ y = 1+ \epsilon$. Where $\epsilon$ is the nil potent matrix with $\epsilon^2 = 0$.

Note:

$$ x^\lambda + y^\lambda = 2 $$ where $\lambda$ can be any integer. $$ xy = 1$$ I believe this structure is telling us about the below $K$ series:

$$ K = sx^2 + s^2x^3 +s^3x^5 + s^4x^7 + s^5 x^{11} + s^6 x^{13}\dots $$ The R.H.S is very similar to:

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

Now we multiply $s$ both sides:

$$ K = 0 + s^2x^2 + s^3x^3 +s^4x^5 + s^5x^7 + s^6 x^{11} +\dots $$ $$ Ks = sx^2 + s^2x^3 +s^3x^5 + s^4x^7 + s^5 x^{11} + s^6 x^{13}+\dots $$

Subtracting both equations:

$$ K(1-s) - sx^2 = s^2x^2(1-x) +s^3x^3(1-x^2) + s^4x^5(1-x^2) + s^5 x^{7}(1-x^4 ) + s^6 x^{11}(1-x^2 ) + \dots $$

Using $ x^\lambda + y^\lambda = 2 $:

$$ K(1-s) - sx^2 = s^2 x^2 (y -1) +s^3x^3(y^2-1) + s^4x^5(y^2-1) + s^5 x^{7}(y^4-1 ) + s^6 x^{11}(y^2-1 ) + \dots $$

We multiply the term within the bracket from the R.H.S and transfer them to the L.H.S and defining them as $K_1$:

$$ K(1-s) + K_1 - sx^2 = s^2 x^2 y +s^3x^3y^2 + s^4x^5y^2 + s^5 x^{7}y^4 + s^6 x^{11}y^2 + \dots $$

Using $xy=1$

$$ K(1-s) + K_1 - sx^2 = s^2 x +s^3x + s^4x^3 + s^5 x^3 + s^6 x^9 + \dots $$

This is similar to: $$ 1,1,3,3,9\dots$$

$\endgroup$
  • $\begingroup$ You say you are taking the absolute values of the differences, but in your later steps, you have negative numbers. If you do not take absolute values, then (calling the original sequence $(p_n^0) = (p_n)$ and the $k$th successor sequence $(p_n^k)$), you have $p_n^k = (-1)^k(F_kp_n - F_{k-1}p_{n+1})$, where $F_k$ is the $k$th Fibonacci number. In particular $p_0^k = (-1)^k(2F_k - 3F_{k-1})$ $\endgroup$ – Paul Sinclair Jan 2 '18 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.