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The prime 127 has 127-1=126 with distinct prime factors 2,3,7 and 127+1=128 with distinct prime factors of only 2; hence 2*3*7=42<127. Log 127/42=q=1.296. Are such primes common? Can a value of q be larger for other primes? I tried before but was not logged in. More examples are 31, with distinct prime factors of 30=2*3*5 and for 32 just 2; hence 2*3*5<31. For 97, 96 has prime factors 2 and 3 and 98 has prime factors 2 and 7; hence 2*3*7=42<97.

[EDIT suggestion to OP: I think the question is asking this: given a prime $p$, calculate the squarefree kernel, $s$, of $p^2-1$ (that is, the product of the distinct prime factors of $p^2-1$); is it common to have $s<p$? is it possible to have $p/s>127/42$? By "common", we may understand "does it happen infinitely often?"] Gerry Myerson understood the question. I can't see why it was closed. Could it be that no one is curious enough to write a program to find more such primes?

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  • $\begingroup$ What does "such" mean? $\endgroup$
    – András Bátkai
    Commented Jun 24, 2016 at 18:25
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    $\begingroup$ I think the question is clear. Given a prime $p$, calculate the squarefree kernel of $p^2-1$ (that is, the product of the distinct prime divisors of $p^2-1$), call it $s$, is it common to have $s<p$? Is it possible to have $p/s>127/42$? $\endgroup$
    – Gerry Myerson
    Commented Jun 25, 2016 at 0:49
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    $\begingroup$ Let $p=577$, then $p-1=576=2^63^2$ and $p+1=578=2\cdot17^2$, so $s=2\cdot3\cdot17=102$, and $p/s=577/102>127/42$. $\endgroup$
    – Gerry Myerson
    Commented Jun 30, 2016 at 3:12
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    $\begingroup$ For $p < 20000000$, the largest value for $q$ is achieved for $p = 4801$, where $s = 210$ and $q \approx 1.58526$. $\endgroup$
    – Stefan Kohl ♦
    Commented Jul 2, 2016 at 22:59
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    $\begingroup$ What about 2101249? Gerhard "Are There Too Many Zeroes?" Paseman, 2016.07.02. $\endgroup$
    – Gerhard Paseman
    Commented Jul 2, 2016 at 23:29

1 Answer 1

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Here is a program written in AWK. You can probably optimize it by aborting the factorization when it is clear that the radical (prod) will be too large to continue. Otherwise you spend a lot of time trial factoring, and it goes slowly once LIM gets above 2000.

BEGIN{  LIM=2000;  LIM2=LIM*LIM;  maxn=maxprod=j=1
   for( n=1; (n++) < LIM2; )  {
      if (n in c) { m=n+(f=c[n]); delete c[n]
           while (m in c) {m+=f};  c[m]=f }
      else {  if (n < LIM) {pr[j++]=n; c[(n+n)]=n }; check(n) }
      }
}

function check(n) { m=n-1 ; o=n+1; prod=1
    for( jj=1; (1<m)||(1<o) ; ) { f=pr[jj++]; add=0
        while (0==m%f) { add=1; m/=f }
        while (0==o%f) { add=1; o/=f }
        if (1==add) prod*=f
        if ((m > 1) && ((f*f) > m)) { prod*=m; m=1 }
        if ((o > 1) && ((f*f) > o)) { prod*=o; o=1 }
        }
    if (n*maxprod > prod*maxn) {maxprod=prod; maxn=n;
        print n, prod, n/(1.0*prod) }
}

I get as output

3 2 1.5
17 6 2.83333
127 42 3.02381
577 102 5.65686
4801 210 22.8619
2101249 23370 89.9122

This strikes me as being related to consecutive smooth numbers m,m+1 such that 2m+1 is prime. You might look up Stormer's theorem on consecutive smooth numbers. I don't know how you will work in the prime bit.

Gerhard "Have Fun With AWK Programs" Paseman, 2016.07.02.

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  • $\begingroup$ The code above may contain errors, as it was hand-entered. You might prefer the implementation at mathoverflow.net/a/50691 which produces prime factor lists. If you write up anything based on either program, I would appreciate attribution. Gerhard "Says Yes To Gift Cards" Paseman, 2016.07.02. $\endgroup$
    – Gerhard Paseman
    Commented Jul 3, 2016 at 0:16
  • $\begingroup$ Perhaps the right question to ask is whether $p/s$ is unbounded. $\endgroup$
    – Gerry Myerson
    Commented Jul 3, 2016 at 2:01

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