A338850 Continued fraction expansion of the smallest constant 'c' such that the numbers 1+floor(c^(n^1.5)) are distinct primes for all n >= 0.

2, 3, 1, 2, 1, 2, 1, 1, 1, 1, 3, 1, 2, 13, 6, 1, 3, 5, 1, 5, 1, 7, 17, 1, 3, 1, 11, 18, 3, 1, 2, 1, 2, 1, 2, 17, 15, 1, 69, 3, 1, 2, 1, 1, 1, 1, 33, 1, 3, 2, 4, 17, 1, 3, 2, 2, 1, 2, 6, 1, 11, 3, 2, 1, 1, 1, 17, 1, 7, 5, 2, 2, 2, 84, 1, 8, 3, 1, 1, 22, 3698, 2, 2, 1, 1, 2, 1, 7, 2, 1, 1, 1, 1, 3, 1, 5, 15, 1, 3, 1, 2, 1, 1, 1, 1, 2, 1, 16, 1, 7, 2, 2, 3, 1, 9

Offset

1,1

Links

Table of n, a(n) for n=1..115.

Example

2+1/(3+1/(1+1/(2+1/(1+1/(2+1/(1+1/(1+1/(1+1/(1+1/(3+1/(1+1/(2+1/(13+1/(6]= 590652/260429 = 2.26799626769... The constant 'c' is equal to 2.267996267706724247328553280725371774527042254400818772275…

Prog

(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);
};
contfrac(c(50, 200), 115)
\\ François Marques, Nov 17 2020

Crossrefs

Cf. A338613, A338837.

Sequence in context: A127246 A106038 A078711 * A322423 A325494 A295561

Adjacent sequences: A338847 A338848 A338849 * A338851 A338852 A338853

Keyword

nonn,cofr

Author

Bernard Montaron, Nov 12 2020

Status

approved