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.

%I A339459 #25 Jul 19 2021 01:23:18
%S A339459 2,7,37,263,2437,28541,414893,7368913,157859813,4035572951,
%T A339459 122006926709,4328504865941,178988464493359,8575347401843113,
%U A339459 473485756611713633,29985730185033339911,2168685169398896331137,178419507110725228550743
%N A339459 Prime numbers a(n) = floor(2^(n^d)) for all n>=1 where d=1.5039285240... is the constant defined at A339457.
%C A339459 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...
%C A339459 Algorithm to generate the smallest constant d and the associated prime number sequence a(n) = floor(2^(n^d)).
%C A339459 0.   n=1, a(1)=2, d=1
%C A339459 1.   n=n+1
%C A339459 2.   b=floor(2^(n^d))
%C A339459 3.   p=smpr(b)     (smallest prime >= b)
%C A339459 4.   If p=b, then a(n)=p, go to 1.
%C A339459 5.   d=log(log(p)/log(2))/log(n)
%C A339459 6.   a(n)=p
%C A339459 7.   k=1
%C A339459 8.   b=floor(2^(k^d))
%C A339459 9.   If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5.
%C A339459 10.  If b is prime then a(k)=b
%C A339459 11.  If k<n-1, then k=k+1, go to 8.
%C A339459 12.  go to 1.
%C A339459 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.
%H A339459 Bernard Montaron, <a href="https://arxiv.org/abs/2011.14653">Exponential prime sequences</a>, arXiv:2011.14653 [math.NT], 2020.
%F A339459 a(n) = floor(2^(n^d)) where d=1.5039285240...
%e A339459 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.
%o A339459 (PARI) A339459(n=30, prec=100) = {
%o A339459 \\ if precision is large enough, returns the list of first n terms of the sequence
%o A339459   my(curprec=default(realprecision));
%o A339459   default(realprecision, max(prec,curprec));
%o A339459   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); );
%o A339459   for(j=1, n-1,
%o A339459     b=floor(c^(j^d));
%o A339459     until(ok,
%o A339459       p=smpr(b);
%o A339459       ok = 1;
%o A339459       listput(a,p,j);
%o A339459       if(p!=b,
%o A339459          d=log(log(p)/log(c))/log(j);
%o A339459          for(k=1,j-2,
%o A339459              b=floor(c^(k^d));
%o A339459              if(b!=a[k],
%o A339459                 ok=0;
%o A339459                 j=k;
%o A339459                 break;
%o A339459                );
%o A339459             );
%o A339459         );
%o A339459     );
%o A339459   );
%o A339459   default(realprecision, curprec);
%o A339459   return(a);
%o A339459 } \\ _François Marques_, Dec 08 2020
%Y A339459 Cf. A339457, A339458, A338613, A338837, A338850.
%K A339459 nonn
%O A339459 1,1
%A A339459 _Bernard Montaron_, Dec 06 2020