cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Original entry on oeis.org

2, 3, 11, 71, 701, 9467, 168599, 3860009, 111498091, 4002608003, 176359202639, 9437436701437, 607818993573569, 46744099128452807, 4262700354254812091, 458091929703695291747, 57691186909930154615407, 8471601990692484416847631, 1443868262009075144775972529
Offset: 0

Views

Author

Bernard Montaron, Nov 03 2020

Keywords

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

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...

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.

Original entry on oeis.org

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

Views

Author

Bernard Montaron, Nov 11 2020

Keywords

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

A339457 Decimal expansion of the smallest positive number d such that numbers of the sequence floor(2^(n^d)) are distinct primes for all n>=1.

Original entry on oeis.org

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

Views

Author

Bernard Montaron, Dec 06 2020

Keywords

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
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.

Examples

			1.5039285240695206335276890678975831991907388495811384290029993506576595475616...

Links

Crossrefs

Cf. A339459, A339458, A338613, A338837, A338850.

Programs

  • 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

A339458 Continued fraction expansion of the smallest constant d such that the numbers floor(2^(n^d)) are distinct primes for all n >= 1.

Original entry on oeis.org

1, 1, 1, 63, 7, 3, 2, 2, 1, 1, 1, 250, 2, 1, 2, 1, 2, 3, 1, 4, 1, 1, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 6, 7, 1, 1, 1, 6, 1, 1, 9, 9, 2, 1, 6, 2, 5, 1, 25, 1, 1, 1, 2, 18, 1, 3, 5, 1, 1, 5, 1, 3, 1, 1, 4, 1, 1, 3, 2, 2, 3, 40, 2, 3, 8, 2, 2, 25, 1, 5, 2, 1, 1, 3, 2, 2, 1, 10, 1, 1, 2, 1, 2, 1, 1, 2, 1, 3, 2, 420, 2, 2, 1
Offset: 1

Views

Author

Bernard Montaron, Dec 06 2020

Keywords

Examples

			1+1/(1+1/(1+1/(63+1/(7+1/(3+1/(2+1/(2+1/(1+1/(1+1/(1+1/(250] = 22739482/15120055 = 1.503928524069522...
The constant is equal to d=1.503928524069520633527689067897583199190738849581138429002999...

Crossrefs

Cf. A339459, A339457, A338613, A338837, A338850.

Programs

  • PARI
    A339458(n=63, prec=200)={
  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;
               );
            );
        );
    );
  );
  default(realprecision, curprec);
  return(contfrac(d));
} \\ François Marques, Dec 08 2020

A339459 Prime numbers a(n) = floor(2^(n^d)) for all n>=1 where d=1.5039285240... is the constant defined at A339457.

Original entry on oeis.org

2, 7, 37, 263, 2437, 28541, 414893, 7368913, 157859813, 4035572951, 122006926709, 4328504865941, 178988464493359, 8575347401843113, 473485756611713633, 29985730185033339911, 2168685169398896331137, 178419507110725228550743
Offset: 1

Views

Author

Bernard Montaron, Dec 06 2020

Keywords

Comments

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...
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
12. go to 1.
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.

Examples

			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.

Links

Crossrefs

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

Programs

  • PARI
    A339459(n=30, prec=100) = {
\\ if precision is large enough, returns the list of first n terms of the sequence
  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;
               );
            );
        );
    );
  );
  default(realprecision, curprec);
  return(a);
} \\ François Marques, Dec 08 2020

Formula

a(n) = floor(2^(n^d)) where d=1.5039285240...
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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