A339459 -id:A339459 - cp's OEIS frontend

A339457 Decimal expansion of the smallest positive number d such that numbers of the sequence floor(2^(n^d)) are distinct primes for all n>=1.

1, 5, 0, 3, 9, 2, 8, 5, 2, 4, 0, 6, 9, 5, 2, 0, 6, 3, 3, 5, 2, 7, 6, 8, 9, 0, 6, 7, 8, 9, 7, 5, 8, 3, 1, 9, 9, 1, 9, 0, 7, 3, 8, 8, 4, 9, 5, 8, 1, 1, 3, 8, 4, 2, 9, 0, 0, 2, 9, 9, 9, 3, 5, 0, 6, 5, 7, 6, 5, 9, 5, 4, 7, 5, 6, 1, 6, 3, 0, 5, 7, 6, 4, 3, 1, 7, 1, 0, 1, 8, 9, 0, 8, 0, 8, 8, 6, 5, 2, 2, 4, 6, 8, 7, 4, 0, 1, 3, 0

Offset: 1

Bernard Montaron, Dec 06 2020

Assuming Cramer's conjecture on prime gaps, it can be proved that there exists at least one constant d such that all floor(2^(n^d)) are primes for n>=1 as large as required. The constant giving the smallest growth rate is d=1.503928524069520633527689067897583199190738...
Algorithm to generate the smallest constant d and the associated prime number sequence a(n)=floor(2^(n^d)).
 n=1, a(1)=2, d=1
 n=n+1
 b=floor(2^(n^d))
 p=smpr(b)     (smallest prime >= b)
 If p=b, then a(n)=p, go to 1.
 d=log(log(p)/log(2))/log(n)
 a(n)=p
 k=1
 b=floor(2^(k^d))
 If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5.
 If b is prime, then a(k)=b
 If k  12.  go to 1.
decimal digits of d are sufficient to calculate the first 50 terms of the prime sequence. The prime number given by the term of index n=50 has 109 decimal digits.

			1.5039285240695206335276890678975831991907388495811384290029993506576595475616...

Bernard Montaron, Exponential prime sequences, arXiv:2011.14653 [math.NT], 2020.

Cf. A339459, A339458, A338613, A338837, A338850.

A339457(n=63, prec=200) = {
\\ returns the list of the first digits of the constant.
\\ the number of digits increases faster than n
  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;
               );
            );
        );
    );
  );
  my(p=floor(-log(d-log(log(a[n-2])/log(c))/log(n-2))/log(10)) );
  default(realprecision, curprec);
  return(digits(floor(d*10^p),10));
} \\ François Marques, Dec 08 2020

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

Original entry on oeis.org

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

Bernard Montaron, Dec 06 2020

			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...

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

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

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A339457 Decimal expansion of the smallest positive number d such that numbers of the sequence floor(2^(n^d)) are distinct primes for all n>=1.

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