A339457 - OEIS

%I #27 Jan 01 2022 09:53:38

%S 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,

%T 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,

%U 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

%N 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.

%C 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...

%C Algorithm to generate the smallest constant d and the associated prime number sequence a(n)=floor(2^(n^d)).

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

%C    1.  n=n+1

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

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

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

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

%C    6.  a(n)=p

%C    7.  k=1

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

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

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

%C   11.  If k<n-1 then k=k+1, go to 8.

%C   12.  go to 1.

%C 112 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.

%H Bernard Montaron, <a href="https://arxiv.org/abs/2011.14653">Exponential prime sequences</a>, arXiv:2011.14653 [math.NT], 2020.

%e 1.5039285240695206335276890678975831991907388495811384290029993506576595475616...

%o (PARI) A339457(n=63, prec=200) = {

%o \\ returns the list of the first digits of the constant.

%o \\ the number of digits increases faster than n

%o   my(curprec=default(realprecision));

%o   default(realprecision, max(prec,curprec));

%o   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   for(j=1, n-1,

%o     b=floor(c^(j^d));

%o     until(ok,

%o       p=smpr(b);

%o       ok = 1;

%o       listput(a,p,j);

%o       if(p!=b,

%o          d=log(log(p)/log(c))/log(j);

%o          for(k=1,j-2,

%o              b=floor(c^(k^d));

%o              if(b!=a[k],

%o                 ok=0;

%o                 j=k;

%o                 break;

%o                );

%o             );

%o         );

%o     );

%o   );

%o   my(p=floor(-log(d-log(log(a[n-2])/log(c))/log(n-2))/log(10)) );

%o   default(realprecision, curprec);

%o   return(digits(floor(d*10^p),10));

%o } \\ _François Marques_, Dec 08 2020

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

%K nonn,cons

%O 1,2

%A _Bernard Montaron_, Dec 06 2020