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 1-3 of 3 results.

A341786 Norms of prime ideals in Z[(1+sqrt(-15))/2], the ring of integers of Q(sqrt(-15)).

Original entry on oeis.org

2, 3, 5, 17, 19, 23, 31, 47, 49, 53, 61, 79, 83, 107, 109, 113, 121, 137, 139, 151, 167, 169, 173, 181, 197, 199, 211, 227, 229, 233, 241, 257, 263, 271, 293, 317, 331, 347, 349, 353, 379, 383, 409, 421, 439, 443, 467, 499, 503, 541, 557, 563, 571, 587
Offset: 1

Views

Author

Jianing Song, Feb 19 2021

Keywords

Comments

The norm of a nonzero ideal I in a ring R is defined as the size of the quotient ring R/I.
Note that Z[(1+sqrt(-15))/2] has class number 2.
Consists of the primes congruent to 1, 2, 3, 4, 5, 8 modulo 15 and the squares of primes congruent to 7, 11, 13, 14 modulo 15.
For primes p == 1, 4 (mod 15), there are two distinct ideals with norm p in Z[(1+sqrt(-15))/2], namely (x + y*(1+sqrt(-15))/2) and (x + y*(1-sqrt(-15))/2), where (x,y) is a solution to x^2 + x*y + 4*y^2 = p; for p == 2, 8 (mod 15), there are also two distinct ideals with norm p, namely (p, x + y*(1+sqrt(-15))/2) and (p, x + y*(1-sqrt(-15))/2), where (x,y) is a solution to x^2 + x*y + 4*y^2 = p^2 with y != 0; (3, sqrt(-15)) and (5, sqrt(-15)) are respectively the unique ideal with norm 3 and 5; for p == 7, 11, 13, 14 (mod 15), (p) is the only ideal with norm p^2.

Examples

			Let |I| be the norm of an ideal I, then:
|(2, (1+sqrt(-15))/2)| = |(2, (1-sqrt(-15))/2)| = 2;
|(3, sqrt(-15))| = 3;
|(5, sqrt(-15))| = 5;
|(17, 7+4*sqrt(-15))| = |(17, 7-4*sqrt(-15))| = 17;
|(2 + sqrt(-15))| = |(2 - sqrt(-15))| = 19;
|(23, 17+4*sqrt(-15))| = |(23, 17-4*sqrt(-15))| = 23;
|(4 + sqrt(-15))| = |(4 - sqrt(-15))| = 31.
		

Crossrefs

The number of distinct ideals with norm n is given by A035175.
Norms of prime ideals in O_K, where K is the quadratic field with discriminant D and O_K be the ring of integers of K: A055673 (D=8), A341783 (D=5), A055664 (D=-3), A055025 (D=-4), A090348 (D=-7), A341784 (D=-8), A341785 (D=-11), this sequence (D=-15*), A341787 (D=-19), A091727 (D=-20*), A341788 (D=-43), A341789 (D=-67), A341790 (D=-163). Here a "*" indicates the cases where O_K is not a unique factorization domain.

Programs

  • PARI
    isA341786(n) = my(disc=-15); (isprime(n) && kronecker(disc,n)>=0) || (issquare(n, &n) && isprime(n) && kronecker(disc,n)==-1)

A296716 Numbers congruent to {7, 11, 13, 29} mod 30.

Original entry on oeis.org

7, 11, 13, 29, 37, 41, 43, 59, 67, 71, 73, 89, 97, 101, 103, 119, 127, 131, 133, 149, 157, 161, 163, 179, 187, 191, 193, 209, 217, 221, 223, 239, 247, 251, 253, 269, 277, 281, 283, 299, 307, 311, 313, 329, 337, 341, 343, 359, 367, 371, 373, 389, 397, 401, 403
Offset: 1

Views

Author

Arkadiusz Wesolowski, Dec 19 2017

Keywords

Comments

For any m >= 0, if F(m) = 2^(2^m) + 1 has a factor of the form b = a(n)*2^k + 1 with k >= m + 2 and n >= 1, then the integer F(m)/b is congruent to 13 or 19 mod 30.

Crossrefs

Programs

  • Magma
    [n: n in [0..403] | n mod 30 in {7, 11, 13, 29}];
    
  • Mathematica
    LinearRecurrence[{1, 0, 0, 1, -1}, {7, 11, 13, 29, 37}, 60]
  • PARI
    Vec(x*(7 + 4*x + 2*x^2 + 16*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2 + O(x^55)))

Formula

a(n) = a(n-1) + a(n-4) - a(n-5), n >= 6.
a(n) = a(n-4) + 30.
G.f.: x*(7 + 4*x + 2*x^2 + 16*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2).
a(n) = (-15 + 5*(-1)^n + (3+9*i)*(-i)^n + (3-9*i)*i^n + 30*n) / 4 where i=sqrt(-1). - Colin Barker, Dec 19 2017
E.g.f.: (5*(e^(-x) + (6*x - 3)*e^x) + 6*cos(x) + 18*sin(x))/4. - Iain Fox, Dec 19 2017

A369863 Inert rational primes in the field Q(sqrt(-21)).

Original entry on oeis.org

13, 29, 43, 47, 53, 59, 61, 67, 73, 79, 83, 97, 113, 127, 131, 137, 149, 151, 157, 163, 167, 181, 197, 211, 227, 229, 233, 241, 251, 281, 311, 313, 317, 331, 349, 379, 383, 389, 397, 401, 409, 419, 433, 449, 463, 467, 479, 487, 499, 503, 547, 557, 563, 569, 571, 577, 587
Offset: 1

Views

Author

Dimitris Cardaris, Feb 03 2024

Keywords

Comments

Primes p such that Legendre(-21,p) = -1.

Crossrefs

Cf. inert rational primes in the imaginary quadratic field Q(sqrt(-d)) for the first squarefree positive integers d: A002145 (1), A003628 (2), A003627 (3), A003626 (5), A191059 (6), A003625 (7), A296925 (10), A191060 (11), A105885 (13), A191061 (14), A191062 (15), A296930 (17), A191063 (19), this sequence (21), A191064 (22), A191065 (23).

Programs

  • Mathematica
    Select[Range[3,600], PrimeQ[#] && JacobiSymbol[-21,#]==-1 &] (* Stefano Spezia, Feb 04 2024 *)
  • SageMath
    [p for p in prime_range(3, 600) if legendre_symbol(-21, p) == -1]
Showing 1-3 of 3 results.