Prove or Disprove the Conjecture on Cousin Primes

Question

Here is the problem in a nutshell -

Ethers or Ether of n, $\mathbf{\eth(n)}$: Ethers of any number n is the set of all the values of every possible permutations of its digits.

Illustration Ethers of $123$, $\eth(123)$= $123, 132, 213, 231, 312$ and $321$. Similarly it follows that $\eth(123)=\eth(132)=\eth(213)=\eth(231)=\eth(312)=\eth(321)$

Magnitude of Ether n, denoted by $|\eth(n)|$ is the total count of Prime Numbers we obtain by considering all of its Ethers individually.

Illustration $|\eth(38)|=\mid\eth(83)|=1$, $|\eth(67)|=1$, $|\eth(251)|=2$, $|\eth(157)|=3$, $|\eth(973)|=4$, $|\eth(2183)|=6$, $|\eth(1379)|=7$, $|\eth(7309)|=9$, $|\eth(5917)|=10$, $|\eth(1237)|=11$ and so on…

---

Zeroes of Ether Any natural number $n$ such that $| \eth(n)|= 0$ is termed as zeroes of Ether, and if such $n$ can be explained, trivially, then it's termed as Trivial Zeroes of Ether.

Trivial Zeroes of Ether are classified as follows

1.) All the multiples of $3$, $| \eth(3m)| = 0$, $m \in{W}$ 2.) The number having any of these digits $(0, 2, 4, 5, 6, 8)$ in any combinations. 3.) Any number $(>11)$ entirely made up of only a single digit $(>1)$. 4.) All the numbers of form $k \cdot 10^\alpha$, $\mathit{| \eth(k\cdot 10^\alpha)| =0, \, 0 \leq k \leq 9, \, \alpha\in{W}}$

---

CONJECTURE 1 If $P_{n+1} - P_n =2$ then there always exists $k$ such that $| \eth(k)| = 0$ for $P_n < k < P_{n+1}$.

CONJECTURE 2 If $P_{n+1} - P_n = 4$ then there exists minimum one $k$ such that $| \eth(k)| \geq 1$ for $P_n < k < P_{n+1}$ where all of its Exceptions are Trivial Zeroes of Ether.

---

It's just a segment from my original post On The Distribution of Prime Numbers I tried my best to handle this Conjecture 2 but couldn't get anywhere. I thought the representation of Primes in base $10$ would incinerate this problem but no progress was made. Any progress will be highly appreciated.

Remark \eth symbol is being utilitized for the new notions of digit permutations.

Edit: Rephrased the Conjecture for better clarification.

Answer 1

CONJECTURE 1: If $P_{n+1}−P_{n}=2$ then there always exists $k$ such that $|\eth(k)|=0$ for $P_n<k<P_{n+1}$

$$$$

(smacks forehead audibly) How could we have missed this ! Oh right ... only one such $k$ exists between them ever It's also a multiple of 3 in all but one instance, making it a trivial case by your definition .

$$$$

CONJECTURE 2 If $P_{n+1}−P_n=4$ then there exists minimum one $k$ such that $|\eth(k)|\geq 1$ for $P_n<k<P_{n+1}$ where all of its Exceptions are Trivial Zeroes of Ether.

Well it can't be their average that's a multiple of 3. A number can only be 2 mod 6, if it contains:

• at least 1 even digit, and the number of 1 mod 3 and digits out numbers the 2 mod 3 digits by a 2 mod 3 amount
• the number of 2 mod 3 digits out number the number of 1 mod 3 digits by a 1 mod 3 amount.

It immediately follows that if the first case is true the number taking off the odd last digit ( 1 or 7 digits rearrangement case) we need it to be 1 mod 3 .

Both cases fail if the digit put last is 2 mod 3 because 5 is the only odd digit that it could be, so it falls to it's companions in 1 or 0 mod 3.

In the case of ending in 0 mod 3, the last digit removal must give 2 mod 3 because if not it isn't in the same equivalence class.

Similar constraints ( though inverted) can be put on in the 4 mod 6 case.

It's still unclear how you can guarantee at least 1.