Bernard Montaron - cp's OEIS frontend

User: Bernard Montaron

Bernard Montaron has authored 8 sequences.

A356435 a(n) is the minimum number of Z x Z lattice points inside or on a circle of radius n^(1/2) for any position of the center of the circle.

0, 2, 4, 8, 10, 14, 16, 20, 22, 26, 29, 32, 32, 39, 41, 44, 46, 51, 52, 56, 58, 62, 66, 69, 69, 74, 79, 82, 85, 88, 88, 92, 96, 100, 103, 106, 108, 113, 116, 119, 120, 122, 124, 132, 135, 138, 141, 143, 145, 146, 152, 158, 160, 164, 164, 166, 172, 175, 179, 181, 184, 186, 189, 193, 194, 199

Offset: 0

Bernard Montaron, Aug 07 2022

nonn

a(n) <= A057655(n).

The terms of square index of this sequence are such that a(n^2) = A123689(2n) >= A291259(n), e.g., a(9) = 26 = A123689(6) >= A291259(3) = 25.

			For n = 1 the minimum number of Z x Z lattice points inside the circle is a(1) = 2. The minimum is obtained, for example, with the circle centered at x = 0.1, y = 0.

Cf. A057655, A123689, A291259.

Let N(u,v,n) be the number of integer solutions (x,y) of (x-u)^2 + (y-v)^2 <= n. Then a(n) is the minimum of N(u,v,n) taken over 0 <= u <= 1/2 and 0 <= v <= u. Due to the symmetries of the square lattice one can limit the position (u,v) of the circle center within this triangle. The terms of the sequence were found by "brute force" search of the minimum of N(u,v,n) for (u,v) running through the triangular domain above.

A356462 a(n) is the maximum number of Z x Z lattice points inside or on a circle of radius n^(1/2) for any position of the center of the circle.

Original entry on oeis.org

1, 5, 9, 12, 14, 21, 21, 24, 28, 32, 37, 37, 41, 45, 48, 52, 52, 57, 61, 63, 69, 69, 72, 76, 78, 81, 89, 89, 92, 97, 97, 100, 104, 112, 112, 115, 116, 121, 122, 127, 129, 137, 137, 140, 144, 148, 148, 152, 155, 157, 161, 164, 169, 177, 177

Offset: 0

Bernard Montaron, Aug 08 2022

nonn

a(n) >= A057655(n).

The terms of square index of this sequence are such that a(n^2) = A123690(2n), e.g., a(9) = 32 = A123690(6).

			For n = 1 the maximum number of Z x Z lattice points inside the circle is a(1) = 5. The maximum is obtained with the circle centered at x = 0, y = 0.

Cf. A057655, A123690, A346993.

Let N(u,v,n) be the number of integer solutions (x,y) of (x-u)^2 + (y-v)^2 <= n. Then a(n) is the maximum of N(u,v,n) taken over 0 <= u <= 1/2 and 0 <= v <= u. The symmetries of the square lattice allow to limit the domain of the circle center (u,v) to this triangle. The terms of this sequence were found by "brute force" search of the maximum of N(u,v,n) for (u,v) in this triangular domain.

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

Bernard Montaron, Dec 06 2020

nonn

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)).
  n=1, a(1)=2, d=1
  n=n+1
  b=floor(2^(n^d))
  p=smpr(b)     (smallest prime >= b)
  If p=b, then a(n)=p, go to 1.
  d=log(log(p)/log(2))/log(n)
  a(n)=p
  k=1
  b=floor(2^(k^d))
  If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5.
 If b is prime then a(k)=b
 If k12.  go to 1.
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.

			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.

Bernard Montaron, Exponential prime sequences, arXiv:2011.14653 [math.NT], 2020.

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

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

a(n) = floor(2^(n^d)) where d=1.5039285240...

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

Bernard Montaron, Dec 06 2020

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)).
 n=1, a(1)=2, d=1
 n=n+1
 b=floor(2^(n^d))
 p=smpr(b)     (smallest prime >= b)
 If p=b, then a(n)=p, go to 1.
 d=log(log(p)/log(2))/log(n)
 a(n)=p
 k=1
 b=floor(2^(k^d))
 If b<>a(k) and b not prime, then p=smpr(b), n=k, go to 5.
 If b is prime, then a(k)=b
 If k  12.  go to 1.
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.

			1.5039285240695206335276890678975831991907388495811384290029993506576595475616...

Bernard Montaron, Exponential prime sequences, arXiv:2011.14653 [math.NT], 2020.

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

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

Bernard Montaron, Dec 06 2020

			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...

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

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

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

Bernard Montaron, Nov 03 2020

nonn

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).
  n=0, a(0)=2, c=2, d=1.5
  n=n+1
  b=1+floor(c^(n^d))
  p=smpr(b)     smallest prime >= b
  If p=b then a(n)=p, go to 1.
  c=(p-1)^(1/n^d)
  a(n)=p
  k=1
  b=1+floor(c^(k^d))
  If b<>a(k) then p=smpr(b), n=k, go to 5.
If k11. 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.

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

Cf. A338837, A338850, A051021, A051254.

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

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

Bernard Montaron, Nov 11 2020

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)).
  n=0, a(0)=2, c=2, d=1.5
  n=n+1
  b=1+floor(c^(n^d))
  p=smpr(b)     smallest prime >= b
  If p=b then a(n)=p, go to 1.
  c=(p-1)^(1/n^d)
  a(n)=p
  k=1
  b=1+floor(c^(k^d))
  If b<>a(k) then p=smpr(b), n=k, go to 5.
If k11. 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.

			2.26799626770672424732855328072537177452704225440081877227559082905078374075...

Cf. A338613, A338850.

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

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.

Original entry on oeis.org

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

Bernard Montaron, Nov 12 2020

			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…

Cf. A338613, A338837.

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

cp's OEIS Frontend

User: Bernard Montaron

A356435 a(n) is the minimum number of Z x Z lattice points inside or on a circle of radius n^(1/2) for any position of the center of the circle.

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A356462 a(n) is the maximum number of Z x Z lattice points inside or on a circle of radius n^(1/2) for any position of the center of the circle.

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

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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.

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

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

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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.

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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.

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