A339458 Continued fraction expansion of the smallest constant d such that the numbers floor(2^(n^d)) are distinct primes for all n >= 1.

1, 1, 1, 63, 7, 3, 2, 2, 1, 1, 1, 250, 2, 1, 2, 1, 2, 3, 1, 4, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 6, 7, 1, 1, 1, 6, 1, 1, 9, 9, 2, 1, 6, 2, 5, 1, 25, 1, 1, 1, 2, 18, 1, 3, 5, 1, 1, 5, 1, 3, 1, 1, 4, 1, 1, 3, 2, 2, 3, 40, 2, 3, 8, 2, 2, 25, 1, 5, 2, 1, 1, 3, 2, 2, 1, 10, 1, 1, 2, 1, 2, 1, 1, 2, 1, 3, 2, 420, 2, 2, 1

Offset: 1

Author

Bernard Montaron, Dec 06 2020

Keywords

• nonn
• cofr

Examples

			1+1/(1+1/(1+1/(63+1/(7+1/(3+1/(2+1/(2+1/(1+1/(1+1/(1+1/(250] = 22739482/15120055 = 1.503928524069522...
The constant is equal to d=1.503928524069520633527689067897583199190738849581138429002999...
		

Crossrefs

Cf. A339459, A339457, A338613, A338837, A338850.

Programs

PARI

A339458(n=63, prec=200)={
  my(curprec=default(realprecision));
  default(realprecision, max(prec,curprec));
  my(a=List([2]), d=1.0, c=2.0, b, p, ok, smpr(b)=my(p=b); while(!isprime(p), p=nextprime(p+1)); return(p); );
  for(j=1, n-1,
    b=floor(c^(j^d));
    until(ok,
      p=smpr(b);
      ok = 1;
      listput(a,p,j);
      if(p!=b,
         d=log(log(p)/log(c))/log(j);
         for(k=1,j-2,
             b=floor(c^(k^d));
             if(b!=a[k],
                ok=0;
                j=k;
                break;
               );
            );
        );
    );
  );
  default(realprecision, curprec);
  return(contfrac(d));
} \\ François Marques, Dec 08 2020