A339457 -id:A339457 - cp's OEIS frontend

A339459 Prime numbers a(n) = floor(2^(n^d)) for all n>=1 where d=1.5039285240... is the constant defined at A339457.

2, 7, 37, 263, 2437, 28541, 414893, 7368913, 157859813, 4035572951, 122006926709, 4328504865941, 178988464493359, 8575347401843113, 473485756611713633, 29985730185033339911, 2168685169398896331137, 178419507110725228550743

Offset: 1

Bernard Montaron, Dec 06 2020

nonn

Assuming Cramer's conjecture on prime gaps is true, it can be proved that there exists at least one constant d such that all terms of the sequence are primes. 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 k12.  go to 1.
decimals 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.

			This example illustrates the importance of doing full precision calculations: a(19) = floor(2^(19^d)) = floor(2^83.7826351429215150692195114432) = 16637432012996855576590853. Here, the precision required on the exponent of 2 is 28 decimals in order to obtain the correct value for a(19). And the precision required keeps increasing with the index value n.

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

Cf. A339457, A339458, A338613, A338837, A338850.

A339459(n=30, prec=100) = {
\\ if precision is large enough, returns the list of first n terms of the sequence
  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(a);
} \\ François Marques, Dec 08 2020

a(n) = floor(2^(n^d)) where d=1.5039285240...

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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A339459 Prime numbers a(n) = floor(2^(n^d)) for all n>=1 where d=1.5039285240... is the constant defined at A339457.

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