A338837 Decimal expansion of the smallest positive number 'c' such that numbers of the form 1+floor(c^(n^1.5)) for all n >= 0 are distinct primes.

2, 2, 6, 7, 9, 9, 6, 2, 6, 7, 7, 0, 6, 7, 2, 4, 2, 4, 7, 3, 2, 8, 5, 5, 3, 2, 8, 0, 7, 2, 5, 3, 7, 1, 7, 7, 4, 5, 2, 7, 0, 4, 2, 2, 5, 4, 4, 0, 0, 8, 1, 8, 7, 7, 2, 2, 7, 5, 5, 9, 0, 8, 2, 9, 0, 5, 0, 7, 8, 3, 7, 4, 0, 7, 5, 1, 4, 6, 9, 5, 7, 3, 5, 7, 2, 1, 7, 3, 8, 3, 6, 2, 9, 0, 9, 9, 2, 5, 7, 0, 0, 4, 2, 7, 3, 1, 5, 8, 7, 3, 1, 7, 1, 1, 5, 7, 6, 5, 8, 8, 1, 9, 3, 4, 0, 9, 7, 2, 8, 1, 1, 3, 9, 0

Offset: 1

Author

Bernard Montaron, Nov 11 2020

Keywords

• nonn
• cons

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...

Algorithm to generate the smallest constant 'c' and the associated prime number sequence a(n)=1+floor(c^(n^1.5)).

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.

The precision of 'c' with the 135 digits listed above is sufficient to calculate the first 50 terms of the prime sequence. The prime number given by the term of index n=49 has 121 decimal digits.

Examples

			2.26799626770672424732855328072537177452704225440081877227559082905078374075...
		

Crossrefs

Cf. A338613, A338850.

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(c);
};
digits(floor(c(55,200)*10^50))
\\ François Marques, Nov 17 2020

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