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

Author

Bernard Montaron, Dec 06 2020

Keywords

• nonn

Comments

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)).

0. n=1, a(1)=2, d=1

1. n=n+1

2. b=floor(2^(n^d))

3. p=smpr(b) (smallest prime >= b)

4. If p=b, then a(n)=p, go to 1.

5. d=log(log(p)/log(2))/log(n)

6. a(n)=p

7. k=1

8. b=floor(2^(k^d))

9. If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5.

10. If b is prime then a(k)=b

11. If k

12. go to 1.

112 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.

Examples

			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.
		

Links

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

Crossrefs

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

Programs

PARI

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

Formula

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

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