2
$\begingroup$

Can every multiple of $3$ be written as arithmetic mean of two pairs of twin prime numbers ???
let's suppose, one of the twin prime pair is $P_1 ,P_1+2$ and another pair is $P_2, P_2+2$. Where $P_1$ and $P_2$ are distinct .
Let's imagine $$T_1=P_1 +(P_1+2)$$ and $$T_2=P_2+(P_2+2)$$ .
$$3n = \frac{T1+T2}{4}$$ where $n\in N$ and $n>2$
For example
$$9 = \frac{(5+7)+(11+13)}{4}$$
Is there any counter example?
It can be also written as $12n=T_1+T_2$
that means every multiple of twelve can be written as summation of two pairs of twin prime numbers

$\endgroup$
  • 2
    $\begingroup$ We don't even know if there are infinitely man pairs of twin primes, so I doubt we can resolve this question in the positive. We might be able to find a counter-example. $\endgroup$ – Thomas Andrews Nov 30 '17 at 14:40
  • $\begingroup$ It is not known that there are infinitely many twin primes. If there are finitely many the answer to your question is certainly no. $\endgroup$ – Mark Bennet Nov 30 '17 at 14:41
  • $\begingroup$ What have you tried? For problems of this form, the first step is generally to check all the cases in some meaningful range. $\endgroup$ – lulu Nov 30 '17 at 14:43
  • $\begingroup$ For that matter, I'm not sure what you mean. Take $12$, the smallest multiple of $12$. Of course $12=5+7$ but you wanted two pairs of twin primes, so how do you get $12$? $\endgroup$ – lulu Nov 30 '17 at 14:47
  • 1
    $\begingroup$ @multipleof3, while the answer to your question is negative, perhaps it only has finitely many counterexamples? See my updated answer. $\endgroup$ – Mees de Vries Nov 30 '17 at 15:05
5
$\begingroup$

There are several counterexamples; written as multiples of 12, examples include 192, 804, and 1032. These were generated by the C++ code below (note: highly inefficient).

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int main() {
    vector<int> primes = {2};
    for(int i = 3; i < (1 << 20); i+=2) {
        for(int p : primes) {
            if(p * p > i) {
                primes.push_back(i);
                break;
            }
            if(i % p == 0)
                break;
        }
    }

    vector<int> twin_primes;
    for(int i = 0; i + 1 < (int)primes.size(); i++) {
        if(primes[i+1] == primes[i] + 2)
            twin_primes.push_back(primes[i]);
    }

    for(int i = 0; i < (1 << 20); i += 12) {
        bool found = false;
        for(int p : twin_primes) {
            for(int q: twin_primes) {
                if(q < p)
                    continue;
                if(i == 2*p + 2*q + 4) {
                    found = true;
                    printf("%d = %d + %d + %d + %d\n", i, p, p+2, q, q+2);
                    break;
                }
            }
            if(found)
                break;
        }
        if(!found) {
            cout << i << " is a counterexample." << endl;
        }
    }
    return 0;
}

Edit: I decided to write a somewhat more efficient version of my code, and found out that up to $2^{26}$ (about 67 million) the only counterexamples, written as multiples of 12 and disregarding 0 and 12, I could find are 192, 804, 1032, 1572, 1812, 2232, 2292, 2532, 2712, 6492, and 8412. So there might still be something to your conjecture; perhaps it only has finitely many counterexamples, and in particular, perhaps 8412 is the largest counterexample?

Edit: it is indeed a standing conjecture that these are the only counterexamples; see the comment by Tito Piezas below. That's surprising! I thought there would be far fewer twin primes than primes.

$\endgroup$
  • 3
    $\begingroup$ +1 Removing the factor of 12, we have, $$16, 67, 86, 131, 151, 186, 191, 211, 226, 541, 701$$ This turns out to be the conjectured finite sequence A243956 $\endgroup$ – Tito Piezas III Nov 30 '17 at 15:22
  • $\begingroup$ @Mees de Vries if there only finitely many counterexample that means it can be true for infintly many number.we can never run out of twin prime! $\endgroup$ – user508245 Nov 30 '17 at 16:01
  • $\begingroup$ @multipleof3: That thought occurred to me as well. However, to prove there are only finitely many counter-examples might be equally as hard a problem. $\endgroup$ – Tito Piezas III Nov 30 '17 at 16:25
  • $\begingroup$ @TitoPiezasIII is there a way to know that they are equally hard? $\endgroup$ – user508245 Nov 30 '17 at 17:13
  • $\begingroup$ @multipleof3: This is just a guess, but if it was more tractable, they would have used this approach to solve the Twin Prime conjecture long ago. $\endgroup$ – Tito Piezas III Dec 1 '17 at 2:15
1
$\begingroup$

(Too long for a comment.)

I. Twin primes $p,\,p+2$

If we define the $11$ integers mentioned above,

$$S=16,67,86,131,151,186,191,211,226,541,701 $$

then,

$$T=6S-2,\; 6S,\; 6S+2$$

are the $33$ integers $N>4$ of A007534,

$$T=94, 96, 98, 400, 402, 404, 514, 516, 518,\dots$$

which is conjectured to be a finite sequence and defined as "Even numbers that are not the sum of a pair of twin primes." (These exceptions are more orderly than the one below.)

II. Cousin primes $p,\,p+4$

If we define the $10+3$ integers,

$$U_1=8, 72, 282, 648, 912, 1062, 1332, 2118, 2298, 2532$$ $$U_2 = 642, 1422, 2952$$ and, $$V_1=U_1-4,\; U_1,\; U_1+4$$ $$V_2=U_2-4,\; U_2$$

then $V_1,V_2$ are the $36$ integers $N>2$ of A133802 which is also conjectured to be a finite sequence and defined as "Even numbers which are not the sum of a pair of cousin primes."

III. Sexy primes $p,\,p+6$

An initial computer search suggests all even $N>8$ is the sum of a pair of sexy primes.

$\endgroup$

Your Answer

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