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 1-6 of 6 results.

A343240 The number of solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)) is given by a(n).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 4, 2, 4, 2, 2, 4, 4, 2, 4, 2, 4, 2, 2, 2, 2, 4, 2, 2, 4, 4, 4, 2, 4, 2, 2, 4
Offset: 1

Views

Author

Wolfdieter Lang, May 16 2021

Keywords

Comments

Row length of irregular triangle A343239.

Examples

			a(19) = 4 because A343238(19) = 42 = 2*3*7 has 2^(1+1) = 4 solutions from the primes 3 and 7.
		

Crossrefs

Programs

  • PARI
    isok(k) = issquare(Mod(-5, k)); \\ A343238
    lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), listput(list, sum(i=0, n-1, Mod(i,n)^2 + 5 == 0)););); Vec(list); \\ Michel Marcus, Sep 17 2023

Formula

a(n) = row length of A343239(n), for n >= 1.
a(1) = a(2) = a(4) = a(8) = 1, and otherwise a(n) = 2^{number of distinct primes from A139513}, that is, primes congruent to {1, 3, 7, 9} (mod 20), appearing in the prime factorization of A343238(n).

A343239 Irregular triangle read by rows giving the solutions x for x^2 == -5 (mod A343238(n)), for x from {0, 1, 2, ..., A343238(n)-1}, for n >= 1.

Original entry on oeis.org

0, 1, 1, 2, 0, 1, 5, 3, 4, 2, 7, 5, 3, 11, 5, 10, 7, 11, 4, 10, 11, 17, 8, 15, 7, 20, 13, 16, 5, 25, 10, 25, 6, 35, 11, 17, 25, 31, 9, 34, 20, 25, 15, 31, 18, 29, 17, 32, 7, 47, 13, 45, 19, 42, 11, 25, 38, 52, 14, 53, 8, 31, 38, 61, 25, 45, 20, 61, 35, 47, 24, 59, 9, 77, 13, 16, 71, 74
Offset: 1

Views

Author

Wolfdieter Lang, May 16 2021

Keywords

Comments

The length of row n is A343240(n).

Examples

			The irregular triangle T with A(n) = A343238(n) begins:
   n  A(n) \ k  1  2  3  4 ...
  ---------------------------
   1,  1:       0
   2,  2:       1
   3,  3:       1  2
   4,  5:       0
   5,  6:       1  5
   6,  7:       3  4
   7,  9:       2  7
   8, 10:       5
   9, 14:       3 11
  10, 15:       5 10
  11, 18:       7 11
  12, 21:       4 10 11 17
  13, 23:       8 15
  14, 27:       7 20
  15, 29:      13 16
  16, 30:       5 25
  17, 35:      10 25
  18, 41:       6 35
  19, 42:      11 17 25 31
  20, 43:       9 34
  ...
		

Crossrefs

Programs

  • PARI
    isok(k) = issquare(Mod(-5, k)); \\ A343238
    lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), my(row = select(x->(Mod(x,n)^2 + 5 == 0), [0..n-1])); listput(list, row))); Vec(list); \\ Michel Marcus, Sep 17 2023

Formula

T(n, k) gives the solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)), for n >= 1.

A344231 Positive integers k properly represented by the positive definite binary quadratic form X^2 + 5*Y^2 = k, in increasing order.

Original entry on oeis.org

1, 5, 6, 9, 14, 21, 29, 30, 41, 45, 46, 49, 54, 61, 69, 70, 81, 86, 89, 94, 101, 105, 109, 126, 129, 134, 141, 145, 149, 161, 166, 174, 181, 189, 201, 205, 206, 214, 229, 230, 241, 245, 246, 249, 254, 261, 269, 270, 281, 294, 301, 305, 309, 321, 326, 329, 334, 345, 349, 366, 369, 381, 389, 401, 405
Offset: 1

Views

Author

Wolfdieter Lang, Jun 10 2021

Keywords

Comments

This is one of the bisections of sequence A343238. The other sequence is A344232.
This is a proper subsequence of A020669.
The primes in this sequence are given in A033205.
Discriminant Disc = -20 = -4*5. Class number h(-20) = A000003(5) = 2. The reduced primitive forms representing the two proper (determinant = +1) equivalence classes are the present principal form F1 = [1, 0, 5] and F2 = [2, 2, 3] treated in A344232.
A positive integer k is properly represented by some primitive form of Disc = -20 if and only if the congruence s^2 + 20 == 0 (mod 4*k) has a solution. See, e.g., Buell Proposition 41, p. 50, or Scholz-Schoeneberg Satz 74, p. 105. That is, x^2 + 5 == 0 (mod k), with s = 2*x. For the representative solutions x from {0, 1, ..., k-1}, with k from A343238, see A343239. These solutions x determine the so-called representative parallel primitive forms (rpapfs) [k, 2*x, (x^2 + 5)/k] representing k. They are properly equivalent (via so called R(t)-transformations) to one of the reduced forms F1 or F2. (See also W. Lang's links in A225953 and A324251, but there indefinite forms are considered.)
In order to find out which k from A343238 is represented either by form F1 or F2 the two generic multiplicative characters of Disc = -20, namely Legendre(k|p), with the odd prime p = 5 which divides Disc = -20, and Jacobi(-1|k) can be used. See Buell, pp. 51-52. They lead to the two classes of genera of Disc -20.
The present genus I, the principal one, has for odd primes p, not 5, the values Legendre(p|5) = Legendre(5|p) = +1 and Jacobi(-1|p) = Legendre(-1|p) = +1, leading for odd primes not equal to 5 to A033205. The prime 2 is not represented. The prime 5 is trivially represented. For the other genus II these two characters have values -1. There prime 2 is represented.
For composite k the prime number factorization is used, and for powers of primes the lifting theorem is employed (see, e.g., Apostol, p. 121, Theorem 5.30). The solution for prime 2 represented by form F2 = [2, 2, 3] (from the other genus II) is not liftable to powers of 2. The solution for prime 5 is also not liftable (proof by induction). The solutions of the other primes from A033205 and A106865 are uniquely liftable to powers of these primes. See A343238 for all properly represented k for Disc = -20.
For the present genus I the properly represented integers k are given by 2^a*5^b*Product_{j=1..PI} (pI_j)^(eI(j))*Product_{k=1..PII} (pII_k)^(eII(k)), with a and b from {0, 1} but if PI = PII = 0 (empty products are 1) then a = b = 0 giving a(1) = 1. The odd primes pI_j are from A033205 (== {1, 9} (mod 20)), the primes pII_k are from the odd primes of A106865 (== {3, 7}(mod 20)). The exponents of the second product are restricted: if a = 1 then PII >= 1 and Sum_{k=1..PII} eII(k) is odd. If a = 0 then PII >= 0, and if PII >= 1 then this sum is even.
Neighboring numbers k (twins) begin: [5, 6], [29, 30], [45, 46], [69, 70], [205, 206], [229, 230], [245, 246], [269, 270], [405, 406], ...
For the solutions (X, Y) of F2 = [1, 0, 5] properly representing k = a(n) see A344233.

References

  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp 121 - 122.
  • D. A. Buell, Binary Quadratic Forms, Springer, 1989.
  • A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, Sammlung Göschen Band 5131, Walter de Gruyter, 1973.

Crossrefs

A344232 All positive integers k properly represented by the positive definite binary quadratic form 2*X^2 + 2*X*Y + 3*Y^2 = k, in increasing order.

Original entry on oeis.org

2, 3, 7, 10, 15, 18, 23, 27, 35, 42, 43, 47, 58, 63, 67, 82, 83, 87, 90, 98, 103, 107, 115, 122, 123, 127, 135, 138, 147, 162, 163, 167, 178, 183, 202, 203, 207, 210, 215, 218, 223, 227, 235, 243, 258, 263, 267, 282, 283, 287, 290, 298, 303, 307, 315, 322, 327, 335, 343, 347, 362, 367, 378, 383, 387
Offset: 1

Views

Author

Wolfdieter Lang, Jun 10 2021

Keywords

Comments

This is one of the bisections of sequence A343238. The other sequence is A344231.
This is a proper subsequence of A029718.
The primes in this sequence are given in A106865.
See A344231 for more details.
The reduced form [2, 2, 3] represents the proper (determinant +1) equivalence class of one of the two genera (genus II) of discriminant -20. The multiplicative generic characters for discriminant Disc = -20 have values Jacobi(a(n)|5) = -1 and Jacobi(-1|a(n)) = -1, for odd a(n) not divisible by 5. See Buell, p. 52.
The product of any two odd a(n), not divisible by 5, is congruent to {1,5} (mod 8). See Buell, 4), p. 51.
For this genus II of Disc = -20 the positive integers represented are given by 2^a*5^b*Product_{j=1..PI} (pI_j)^(eI(j))*Product_{k=1..PII}(pII_k)^(eII(k)), with a and b from {0, 1}, but if PI = PII = 0 (empty products are 1) then (a, b) = (1, 0) or (1, 1), giving a(1) = 2 or a(4) = 10. The odd primes pI_j are from A033205 and the odd primes pII_j from the odd primes of A106865. The exponents of the second product satisfy: if a = 1 then PII >= 0, and if PII >=1 then Sum_{k=1..PII} eII(j) is even. If a = 0 then PII >= 1 and this sum is odd.
The neighboring numbers k (twins) begin: [42, 43], [82, 83], [122, 123] [162, 163], [202, 203], [282, 283], ...
For the solutions (X, Y) of F2 = [2, 2, 3] properly representing k = a(n) see A344234.

References

  • D. A. Buell, Binary Quadratic Forms, Springer, 1989.
  • A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, Sammlung Göschen Band 5131, Walter de Gruyter, 1973.

Crossrefs

A344234 Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation 2*X^2 + 2*X*Y + 3*Y^2 = A344232(n), for n >= 1.

Original entry on oeis.org

1, 0, 0, 1, 1, -1, 1, 1, 2, -1, 1, -2, 2, 1, 3, -1, 1, 2, 3, -2, 1, -3, 2, -3, 3, 1, 4, -1, 1, 3, 4, -3, 1, -4, 3, 2, 3, -4, 5, -2, 4, 1, 5, -1, 5, 2, 7, -2, 4, 3, 7, -3, 1, 5, 6, 1, 6, -5, 7, -1, 3, 4, 7, -4, 1, -6, 5, -6
Offset: 1

Views

Author

Wolfdieter Lang, May 17 2021

Keywords

Comments

The length of row n is r(n) = 2*A343240(b(n)), if A344232(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 2, 2, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, ...).
See A344231 for references and links on parallel forms and half-reduced right neighbor forms (R-transformations), and also for the remark on the equivalent reduced form [2, -2, 3].
The number of proper solutions (X, Y), with X > 0, is 1 for n = 1 and 4. X = 0 only for n = 2, but the solution (0, -1) = (-0, -1) is not listed here.
For other n each distinct odd prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, that divides A344232(n) contributes a factor of 2 to the listed number of solutions. See A343238 and A343240 for the multiplicities.
Only solutions with nonnegative X are listed. There is also the corresponding solution (-X, -Y). Hence the total number of signed solution is twice the number considered here.

Examples

			The irregular triangle T(n, m) begins (A(n) = A344232(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   2:       1  0
2,   3:       0  1   1 -1
3,   7:       1  1   2 -1
4,  10:       1 -2
5,  15:       2  1   3 -1
6,  18:       1  2   3 -2
7,  23:       1 -3   2 -3
8,  27:       3  1   4 -1
9,  35:       1  3   4 -3
10, 42:       1 -4   3  2   3 -4  5 -2
11, 43:       4  1   5 -1
12, 47:       2  3   5 -3
13, 58:       1  4   5 -4
14, 63:       2 -5   3 -5   5  1   6 -1
15, 67:       1 -5   4 -5
16, 82:       5  2   7 -2
17, 83:       4  3   7 -3
18, 87:       1  5   6  1   6 -5   7 -1
19, 90:       3  4   7 -4
20, 98:       1 -6   5 -6
...
n = 2: The prime 3 is a member of A139513, hence 2^1 = 2 solutions are listed. There are also the corresponding (-X, -Y) solutions.
n = 4: 10 = A344232(4) = A343238(8) = 2*5, A343240(8) = 1, hence there is 1 pair of proper solution with X >= 0. This is because neither 2 nor 5  are primes from A139513. There is also the solution (-1, 2).
n = 6: Prime 3 is a member of A139513, not prime 2. This there are 2 solutions listed. The solution (3, 0) does not appear; it is not proper.
n = 10: 42 = A344232(10) = A343238(19) = 2*3*7,  A343240(19) = 2^2 = 4, hence there are 4 pairs of proper solution with X >= 0. 3 and 7 are primes from A139513.
		

Crossrefs

Formula

T(n, m) gives for m = 2^j - 1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of 2*T(n, 2*j-1)^2 + 2*T(n, 2*j-1)*T(n, 2*j) + 3*T(n, 2*j)^2 = A344232(n), for j = 1, 2, ..., r(n), for n >= 1. For n = 2 the solution (0, -1) is not listed here.

A344233 Irregular triangle read by rows: row n gives the pairs of proper solutions (X, Y), with gcd(X, Y) = 1 and X >= 0, of the Diophantine equation X^2 + 5*Y^2 = A344231(n), for n >= 1.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, May 17 2021

Keywords

Comments

The length of row n is r(n) = 2*A343240(b(n)), if A344231(n) = A343238(b(n)), for n >= 1. This sequence begins 2*(1, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, ...).
The number of proper solutions (X, Y), with X > 0, is 1 for n = 1. X = 0 only for n = 2, and for n >= 2 only one half of the solutions are listed, namely those with X >= 0. There are also the solutions with (-X, -Y). Thus the total number of solutions for n >= 2 is actually r(n) given above.
For n >= 2 each distinct odd primes from {1, 3, 7, 9} (mod 20), i.e., from
A139513, that divides A344231(n) contributes a factor 2 to the number of solutions listed here. See A343238 and the corresponding A343240.

Examples

			The irregular triangle T(n, m) begins (A(n) = A344231(n)):
n   A(n) \ m  1  2   3  4   5  6   7  8 ...
1,   1:       1  0
2,   5:       0  1
3,   6:       1  1   1 -1
4,   9:       2  1   2 -1
5,  14:       3  1   3 -1
6,  21:       1  2   1 -2   4  1   4 -1
7,  29:       3  2   3 -2
8,  30:       5  1   5 -1
9,  41:       6  1   6 -1
10, 45:       5  2   5 -2
11, 46:       1  3   1 -3
12, 49:       2  3   2 -3
13, 54:       7  1   7 -1
14, 61:       4  3   3 -3
15, 69:       8  1   8 -1   7  2   7 -2
16, 70:       5  3   5 -3
17, 81:       1  4   1 -4
18, 86:       9  1   9 -1
19, 89:       3  4   3 -4
20, 94:       7  3   7 -3
...
n = 2: Prime 5 is not a member of A139513, therefore only 1 solution appears here (see the remark above on the solution (0, -1)).
n = 4: Prime 3 a member A139513. Thus there are 2^1 = 2 solutions are listed. The solution (3, 0) does not appear; it is not proper.
n = 6: 21 = A344231(6) = A343238(12) = 3*7. hence A343240(12) = 2^2 = 4 and there are 4 pairs of proper solutions with X >= 0.
		

Crossrefs

Formula

T(n, m) gives for m = 2^j-1, the nonnegative X(n, j) solution, and for m = 2*j the Y(n, j) solution of T(n, 2*j-1)^2 + 5*T(n, 2*j)^2 = A344231(n), for j = 1 ..r(n), for n >= 1. For n = 2 (X(2) = 0) the solution (0, -1) is not listed.
Showing 1-6 of 6 results.