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A quick check shows this to be true for odd numbers less than 3289. Is this a known result?

Here is the PARI code I used to check:

x=[]; forprime(p=3, 300, forprime(q=3, 300, forprime(r=3, 300, if(2*p*q-r>0 && p!=q && p!=r && q!=r, x=setunion(x, [2*p*q-r])) )));for(i=1, length(x), if(x[i]<10000,print(x[i])))

The first odd number this particular check fails for is $3289$, but $3289=(2)(5)(347)-181$.

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  • $\begingroup$ Your question answers the question you asked in the title. As to whether this result is new, probably yes. Why do you care? It's easy to make up questions like this that fail somewhere easy to find. Did this one come from an interesting place? $\endgroup$ Nov 25 '19 at 16:41
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    $\begingroup$ @EthanBolker The conjecture doesn't fail for $3289$. It's just that the program fails for $3289$. In fact $3289 = 2 \cdot 3 \cdot 557 - 53$. $\endgroup$ Nov 25 '19 at 16:43
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    $\begingroup$ I wrote a C# program and it checked up to $10^9$, and all worked. The one with the largest $r$ was $214848253=2*3*35808151-653$. $\endgroup$ Nov 25 '19 at 16:57
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    $\begingroup$ It would make a good OEIS sequence: let $a(n)$ be the least prime $r$ such that $2n+1+r = 2pq$ for odd primes $p$ and $q$. $\endgroup$ Nov 25 '19 at 17:47
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    $\begingroup$ Contributed to OEIS as sequence A329951. $\endgroup$ Nov 25 '19 at 20:06
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It's very likely that every odd $n$ can be represented this way. In fact, let $p$ be any odd prime that does not divide $n$. Then Dickson's conjecture implies there are infinitely many $q$ such that $q$ and $2pq - n$ are both prime.

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    $\begingroup$ +1 Nice to see yet another consequence of Dickson's Conjecture. $\endgroup$ Nov 25 '19 at 16:46
  • $\begingroup$ @RobertIsrael To finish the thought, if we let $r=2pq-n$ then we have $2pq-r=2pq-2pq-n=n$. So, Dickson's Conjecture not only implies each odd number has a representation, there are infinitely many? $\endgroup$
    – Goldbug
    Nov 25 '19 at 17:17
  • $\begingroup$ $$-s + ((4\cdot u + 2)\cdot t + 2\cdot u)$$$$s,t,u\notin \{v:v=2\cdot x\cdot y+x+y\}$$ should be the form of all natural numbers then. $\endgroup$
    – user645636
    Nov 25 '19 at 20:53

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