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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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%I #25 Jul 19 2021 01:23:18

%S 2,7,37,263,2437,28541,414893,7368913,157859813,4035572951,

%T 122006926709,4328504865941,178988464493359,8575347401843113,

%U 473485756611713633,29985730185033339911,2168685169398896331137,178419507110725228550743

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

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

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

%e 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 (PARI) A339459(n=30, prec=100) = {

%o \\ if precision is large enough, returns the list of first n terms of the sequence

%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 default(realprecision, curprec);

%o return(a);

%o } \\ _François Marques_, Dec 08 2020

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

%K nonn

%O 1,1

%A _Bernard Montaron_, Dec 06 2020