A338613 Numbers given by a(n) = 1 + floor(c^(n^1.5)) where c=2.2679962677... is the constant defined at A338837.

2, 3, 11, 71, 701, 9467, 168599, 3860009, 111498091, 4002608003, 176359202639, 9437436701437, 607818993573569, 46744099128452807, 4262700354254812091, 458091929703695291747, 57691186909930154615407, 8471601990692484416847631, 1443868262009075144775972529

Offset: 0

Author

Bernard Montaron, Nov 03 2020

Keywords

• nonn

Comments

Assuming Cramer's conjecture on largest prime gaps, it can be proved that there exists at least one constant 'c' such that all a(n) are primes for n as large as required. The constant giving the smallest growth rate is c=2.2679962677067242473285532807253717745270422544...

This exponential sequence of prime numbers grows very slowly compared to Mills' sequence for which each new term has 3 times more digits than the previous one. More than 60 terms (all prime numbers) can be easily calculated for the sequence described here which is quite remarkable for an exponential sequence.

Algorithm to compute the smallest constant 'c' and the associated prime number sequence a(n).

0. n=0, a(0)=2, c=2, d=1.5

1. n=n+1

2. b=1+floor(c^(n^d))

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

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

5. c=(p-1)^(1/n^d)

6. a(n)=p

7. k=1

8. b=1+floor(c^(k^d))

9. If b<>a(k) then p=smpr(b), n=k, go to 5.

10. If k

11. go to 1.

I propose the following generalization: find the function f(n) with f(0)=0 and f(x)>x for x>=2 such that there exists a suitable positive constant c(f) giving the increasing prime sequence a(n)=1+floor(c^f(n)) with the smallest possible growth rate. Since a(0)=2, c(f)>=2.

Links

• François Marques, Table of n, a(n) for n = 0..199
• Bernard Montaron, Exponential prime sequences, arXiv:2011.14653 [math.NT], 2020.

Crossrefs

Cf. A338837, A338850, A051021, A051254.

Programs

PARI

c(n=40, prec=100)={
  my(curprec=default(realprecision));
  default(realprecision, max(prec, curprec));
  my(a=List([2]), d=1.5, 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=1+floor(c^(j^d));
    until(ok,
      ok=1;
      p=smpr(b);
      listput(a,p,j+1);
      if(p!=b,
         c=(p-1)^(j^(-d));
         for(k=1,j-2,
             b=1+floor(c^(k^d));
             if(b!=a[k+1],
                ok=0;
                j=k;
                break;
               );
            );
        );
    );
  );
  default(realprecision, curprec);
  return(a);
} \\ François Marques, Nov 12 2020

Formula

a(n) = 1 + floor(c^(n^1.5)) where c=2.2679962677...

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