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We define recursively $p_1=2,p_2=3$ and $$p_{n}= \min_{(A,B)\in F_{n-1}}|A-B| $$ Where $$ \begin{split} F_n=\{(A,B) |&\gcd (A,B)=1,\quad |A-B| \not =1, \\\ &\text{both $A$ and $B$ are products of powers of $p_i$ for $i\le n$}, \\\ &\text{for each $i\le n$, either $p_i |A$ or $p_i |B$}\} \end{split} $$

Is always $p_n$ the $n-th$ prime?

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  • $\begingroup$ Could you please explain more verbally (or by a correct formula) how $F_{n-1}$ is defined. Am I correct to assume that both $A$ and $B$ are divisible by (at least) one of the $p_i$ but not nececessarily the same one. And that the second use of $(A,B)$ means the gcd, while the first means the couple. $\endgroup$
    – user9072
    Commented Mar 9, 2013 at 12:18
  • $\begingroup$ @quid yes i wanted to say that only $p_i$ divide A and B,but i think that it doesnt affect the definition of minimum,thus $p_n$ is the same. $\endgroup$
    – Asterios Gkantzounis
    Commented Mar 9, 2013 at 12:30
  • $\begingroup$ Sorry, it is still not clear to me what you are trying to say. What I now assume is that $A$ and $B$ are both products of $p_1, \dots, p_{n-1}$ (allowing repetead factors). But then how should this ever work? Assuming $p_3$ will be indeed $5$ given by $(4,9)$. But then how is the condition to be read that in the next step $(3,5)$ is excluded for a minimum of $2$? $\endgroup$
    – user9072
    Commented Mar 9, 2013 at 12:56
  • $\begingroup$ $p_4=7=2^3-1$, is it legal in your definition ? do you admit the zeroth power ? $\endgroup$
    – Duchamp Gérard H. E.
    Commented Mar 9, 2013 at 13:05
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    $\begingroup$ It seems that there are two questions here, the one in the title and the one in the body, which are not equivalent. $\endgroup$
    – François Brunault
    Commented Mar 9, 2013 at 21:53

2 Answers 2

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Barry Cipra has already computed the first few values. The next couple of numbers $p_n$ are

$13 = 5 \cdot 11 - 2 \cdot 3 \cdot 7$,

$17 = 2 \cdot 7 \cdot 13 - 3 \cdot 5 \cdot 11$,

$19 = 2^2 \cdot 3 \cdot 5 \cdot 17 - 7 \cdot 11 \cdot 13 $,

$23 = 7 \cdot 11 \cdot 17^2 - 2 \cdot 3^2 \cdot 5 \cdot 13 \cdot 19$,

$29 = 3 \cdot 11 \cdot 13^2 \cdot 19 \cdot 23 - 2^{12} \cdot 5 \cdot 7 \cdot 17$.

I don't find such expression for 31 in numbers $\leq 10^{12}$.

The smallest I find is $47 = 3 \cdot 7 \cdot 19^2 \cdot 23 \cdot 29 - 2^5 \cdot 5 \cdot 11 \cdot 13^2 \cdot 17$.

If the abc conjecture is true, there are at most finitely many ways to express the $n$-th prime as a difference of coprime numbers which are divisible only by the first $n-1$ primes.

Addendum: For a more extensive table, see http://www.fermatquotient.com/DiverseMinimas/S=M-N_min (found by Google'ing for the numbers obtained above). The data supports the assumption that the OP's assertion is unlikely to be true.

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  • $\begingroup$ Given that the OP also requires the product of the two coprime numbers $A$ and $B$ to be divisible by all of the first $n-1$ $p_i$s, is it reasonable to infer from what the $abc$ conjecture says that there are likely to be counterexamples (with 31 possibly the first)? $\endgroup$
    – Barry Cipra
    Commented Mar 9, 2013 at 16:32
  • $\begingroup$ @Barry Cipra: at first glance, for the ABC reasoning this does not help that much as the upper-bound like this is worst possible in this case (of course it limits the choices under this bound, bt still). In general to me it seems ABC would (only) allow to (better) quantifiy the heuristic that since the numbers that are products of a fixed set of primes are (eventually) quite sparse there is no reason to expect any of them being close together "later on". And it is in fact know unconditioannly that there can be at most most finitely many solutions for every fixed difference... $\endgroup$
    – user9072
    Commented Mar 9, 2013 at 17:18
  • $\begingroup$ ...and in that sense I would say if one does not find the 31 somewhat soon, one will "likely" never find it. What I do not know is wheter it is possible to make the bounds one can get good enough to make a check up to these bounds feasible. $\endgroup$
    – user9072
    Commented Mar 9, 2013 at 17:21
  • $\begingroup$ It's quite plausible that there are heuristics suggesting the existence of counterexamples. -- But the abc conjecture doesn't tell very much directly: what we have is $a+b=c$ where $a,b,c$ are coprime and the squarefree part $q(abc)$ of $abc$ equals the product of the first $n$ primes. Now the abc conjecture asserts that for any $\epsilon > 0$ there is a constant $C_{\epsilon}$ such that $c < C_{\epsilon} \cdot q(abc)^{1+\epsilon}$. Since not all of the first $n$ primes divide $c$ (in general maybe only something like half of them), there remains still some room to vary exponents of powers. $\endgroup$
    – Stefan Kohl ♦
    Commented Mar 9, 2013 at 17:22
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You can find some amusing (I might say amazing) papers in this area by searching for "primes at a glance" and " primes at a (somewhat lengthy) glance".

In the first paper they give (along with many other interesting things) what they "believe to be a complete list" of all pairs of integers $B,L$ with

  1. $N=B+L$
  2. $B \ge |L|$
  3. $\gcd(B,L)=1$
  4. if $p \le\sqrt{N}$, then $p$ divides $BL.$
  5. if $Q | BL$ and $p \lt q$ then $p | BL$

For $N \gt 1$ this proves $N$ to be prime as it rules out any proper divisors. Such a presentation provides an at a glance proof that $p$ is prime. For $N=31$ the presentations range from

$31=2^4+3\cdot5$

to

$31=2^3\cdot7\cdot11\cdot17\cdot23-3\cdot5^2\cdot13^2\cdot19$

The first primes for which they give no solutions are $541,547$

the last for which they do is

$2521=19\cdot43\cdot37\cdot2\cdot3^2\cdot5\cdot29^2\cdot41\cdot47^2-7\cdot11^3\cdot13\cdot17^2\cdot23^2\cdot31$

Without condition 5 there are many solutions. $88711$ is the product of $7,19,23,29$ and $72930$ is the product of $2,3,5,11,13,17$ so we can certainly find positive coprime integers with $1=88711x-72930y.$ Then $31=88711s-72930t$ is a difference of coprime values for $s=31x+72930$ and $t=31y+88711$ You can always do that.But probably not with $st$ having all prime factors below 31 ( in which case the prime factors would split the same way, given that none of them divide $31$.)

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    $\begingroup$ I do not fully understand your last paragraph, as this does not seem equivalent to the question asked: the primes could "distribute" differently over the $s$ and $t$ than over the "coefficients." $\endgroup$
    – user9072
    Commented Mar 9, 2013 at 22:24
  • $\begingroup$ That could only happen if one of them divided 31. But I added an explanation. $\endgroup$
    – Aaron Meyerowitz
    Commented Mar 10, 2013 at 4:07
  • $\begingroup$ Thank you for the reply and the added explanation! And, sorry for the noise, I should have realized this myself. $\endgroup$
    – user9072
    Commented Mar 10, 2013 at 12:07

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