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.

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A341790 Norms of prime elements in Z[(1+sqrt(-163))/2], the ring of integers of Q(sqrt(-163)).

Original entry on oeis.org

4, 9, 25, 41, 43, 47, 49, 53, 61, 71, 83, 97, 113, 121, 131, 151, 163, 167, 169, 173, 179, 197, 199, 223, 227, 251, 263, 281, 289, 307, 313, 347, 359, 361, 367, 373, 379, 383, 397, 409, 419, 421, 439, 457, 461, 487, 499, 503, 523, 529, 547, 563, 577, 593
Offset: 1

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Author

Jianing Song, Feb 19 2021

Keywords

Comments

Also norms of prime ideals in Z[(1+sqrt(-163))/2], which is a unique factorization domain. The norm of a nonzero ideal I in a ring R is defined as the size of the quotient ring R/I.
Consists of the primes such that (p,163) >= 0 and the squares of primes such that (p,163) = -1, where (p,163) is the Legendre symbol.
For primes p such that (p,163) = 1, there are two distinct ideals with norm p in Z[(1+sqrt(-163))/2], namely (x + y*(1+sqrt(-163))/2) and (x + y*(1-sqrt(-163))/2), where (x,y) is a solution to x^2 + x*y + 41*y^2 = p; for p = 163, (sqrt(-163)) is the unique ideal with norm p; for primes p with (p,163) = -1, (p) is the only ideal with norm p^2.

Examples

			norm((1 + sqrt(-163))/2) = norm((1 - sqrt(-163))/2) = 41;
norm((3 + sqrt(-163))/2) = norm((3 - sqrt(-163))/2) = 43;
norm((5 + sqrt(-163))/2) = norm((5 - sqrt(-163))/2) = 47;
norm((7 + sqrt(-163))/2) = norm((7 - sqrt(-163))/2) = 53;
...
norm((79 + sqrt(-163))/2) = norm((79 - sqrt(-163))/2) = 1601.
		

Crossrefs

The number of nonassociative elements with norm n (also the number of distinct ideals with norm n) is given by A318983.
The total number of elements with norm n is given by A318985.
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), A341786 (D=-15*), A341787 (D=-19), A091727 (D=-20*), A341788 (D=-43), A341789 (D=-67), this sequence (D=-163). Here a "*" indicates the cases where O_K is not a unique factorization domain.

Programs

  • PARI
    isA341783(n) = my(disc=-163); (isprime(n) && kronecker(disc,n)>=0) || (issquare(n, &n) && isprime(n) && kronecker(disc,n)==-1)

A296921 Rational primes that decompose in the field Q(sqrt(-163)).

Original entry on oeis.org

41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 167, 173, 179, 197, 199, 223, 227, 251, 263, 281, 307, 313, 347, 359, 367, 373, 379, 383, 397, 409, 419, 421, 439, 457, 461, 487, 499, 503, 523, 547, 563, 577, 593, 607, 641, 647, 653, 661, 673, 677, 691, 701, 709, 733, 739, 743, 773, 787, 797
Offset: 1

Views

Author

N. J. A. Sloane, Dec 25 2017

Keywords

Comments

From Jianing Song, Oct 13 2022: (Start)
Primes p such that kronecker(-163,p) = 1 (or equivalently, kronecker(p,163) = 1).
Primes p such that p^81 == 1 (mod 163). (End)

Crossrefs

A257362, the sequence of primes that do not remain inert in the field Q(sqrt(-163)), is essentially the same.
Cf. A296915 (rational primes that remain inert in the field Q(sqrt(-163))).

Programs

  • Maple
    Load the Maple program HH given in A296920. Then run HH(-163,200);
  • PARI
    isA296921(p) = isprime(p) && kronecker(p,163) == 1

A330673 The possible v-factors for any A202018(k) (while A202018(k) = v * w, v and w are integers, w >= v >= 41, v = w iff w = 41, all such v-factors form the set V).

Original entry on oeis.org

41, 43, 47, 53, 61, 71, 83, 97, 113, 131, 151, 163, 167, 173, 179, 197, 199, 223, 227, 251, 263, 281, 307, 313, 347, 359, 367, 373, 379, 383, 397, 409, 419, 421, 439, 457, 461, 487, 499, 503, 523, 547, 563, 577, 593, 607, 641, 647, 653, 661, 673, 677, 691, 701, 709, 733
Offset: 0

Views

Author

Sergey Pavlov, Dec 23 2019

Keywords

Comments

This is different from A257362: a(n) = A257362(n+1) for n=0..109, but a(110) = 1468 != 1471 = A257362(111). - Alois P. Heinz, Mar 02 2020
A kind of prime number sieve for the numbers of form x^2+x+41 (for so-called Euler primes, or A005846).
A set of all composite Euler numbers of form x^2+x+41 could be written as a 4-dimensional matrix m(i,j,t,u); a set of all terms of a(n) could be written as a 3-dimensional matrix v(i,j,t), since, for any integer u > -1, and for any w-factor that has the same values for i, j, t, we have the same v-factor (u = -1 iff w = 41); see formulas below.
Theorem. Let m be a term of A202018. Then m is composite iff m == 0 (mod v), where v is a term of a(n), v <= sqrt(m) (v = sqrt(m) iff m = 1681); otherwise, m is prime. Moreover, while m == 0 (mod p) (p is prime, p <= sqrt(m), p = sqrt(m) iff m = 1681), p is a term of a(n).
While i = 1, any v(i,t,j) is a term of both A202018 and a(n) (trivial).
Any w is a term of V and of a(n) which is the superset of V.

Examples

			Let  i = 3, t = 1, j = -1. Then v(i,t,j) = m(j) * i^2 + b + ja = 41 * 3^2 + 4 - 6 = 41 * 9 - 2 = 367, and 367 is a term of a(n).
We could find all terms of a(n) v < 10^n and then all Euler primes p < 10^(2n) (for n > 1, number of all numbers m such that are terms of A202018 (and any m < 10^(2n)) is 10^n; trivial).
Let 2n = 10; it's easy to establish that, while i > 49, any v(i,t,j)^2 > 10^10; thus, we can use 0 < i < 50 to find all numbers v < 10^5. While m is a term of A202018, m < 10^10, m is composite iff there is at least one v such that m == 0 (mod v); otherwise, m is prime. We could easily remove all "false" numbers v that cannot be divisors of any m. Let p' be a regular prime (p' is a term of A000040, but not of a(n)) such that any 3p' < UB(i); in our case, any 3p' < 50. Thus, we could try any v with p' = {2,3,5,7,11,13}; if v == 0 (mod p'), it is "false"; otherwise, there is at least one m < 10^10 such that m == 0 (mod v).
		

Crossrefs

Formula

Let j = {-1;0;-2;1;-3;2;...;-(n+1);n}, m(-1) = 41, m(0) = 41, etc. (while j is negative, m(j) = A202018(-(j+1)); while j is nonnegative, m(j) = A202018(j)). Any term of a(n) could be written at least once as v(i,t+1,j) = m(j) * i^2 + b + ja, where i, t, and j are integers (j could be negative), i > 2; a = (i^2 - 2i) - 2i(t - 1), b = a - ((i^2 - 4)/4 - ((t - 1)^2 + 2(t - 1))), 0 < t < (i/2), while i is even; a = (i^2 - i) - 2i(t - 1), b = a - ((i^2 - 1)/4 - ((t - 1)^2 + (t - 1))), 0 < t < ((i + 1)/2), while i is odd (Note: v(i,1,j) = v(i,i/2,j), while i is even; v(i,1,j) = v(i,(i + 1)/2,j), while i is odd); at i = 2, v(2,1,j) = 4 * m(j) + 3 + 4j (at i = 2, we use only j < 0); at i = 1, v(1,1,j) = m(j) (at i = 1, we use only j >= 0; trivial).
Showing 1-3 of 3 results.