OFFSET
1,2
COMMENTS
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...
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<n-1 then k=k+1, go to 8.
12. go to 1.
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.
LINKS
Bernard Montaron, Exponential prime sequences, arXiv:2011.14653 [math.NT], 2020.
EXAMPLE
1.5039285240695206335276890678975831991907388495811384290029993506576595475616...
PROG
(PARI) A339457(n=63, prec=200) = {
\\ returns the list of the first digits of the constant.
\\ the number of digits increases faster than n
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;
);
);
);
);
);
my(p=floor(-log(d-log(log(a[n-2])/log(c))/log(n-2))/log(10)) );
default(realprecision, curprec);
return(digits(floor(d*10^p), 10));
} \\ François Marques, Dec 08 2020
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Bernard Montaron, Dec 06 2020
STATUS
approved