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.

A232550 Number of distinct primitive quadratic forms of discriminant = -4n that exist such that every prime p for which p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n) can be represented by one of them.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 3, 1, 2, 2, 1, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 5, 1, 2, 3, 2, 2, 2, 2, 2, 3, 2, 1, 4, 1, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 5, 2, 1, 3, 2, 1, 4, 2, 2, 2, 1, 3, 3, 2, 2, 3, 2, 2
Offset: 1

Views

Author

V. Raman, Nov 26 2013

Keywords

Comments

A quadratic form is primitive if the GCD of the coefficients is 1. For example, the quadratic form 2*x^2+4*y^2 is not primitive.
Two quadratic forms f(x,y) = a*x^2+b*x*y+c*y^2 and g(x,y) = p*x^2+q*x*y+r*y^2 are distinct (or inequivalent) if and only if one cannot be obtained by a linear transformation (of the variables x, y) from the other. For example, the three quadratic forms u(x,y) = 3*x^2+2*x*y+3*y^2, v(x,y) = 3*x^2+4*x*y+4*y^2 and w(x,y) = 3*x^2+10*x*y+11*y^2 are equivalent because v(x,y) = u(x+y,-y) and w(x,y) = v(x+y,y). Also, w(x,y) = u(x+2*y,-y). Similarly, the two quadratic forms s(x,y) = 8*x^2+9*y^2 and t(x,y) = 17*x^2+50*x*y+41*y^2 are equivalent because t(x,y) = s(x+2*y,x+y).
The quadratic form x^2+n*y^2 is one such form and the only such form if n is a convenient number (A000926).
a(n) = 1 if and only if n is a convenient number (A000926).
Any prime p such that p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n) can be represented by exactly one of the a(n) distinct primitive quadratic forms of discriminant = -4n in at most four different ways (if n >= 2) or in at most eight different ways (if n = 1).
If a prime p can be written in the form x^2+n*y^2, then either p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), assuming that p^2 does not divide n.
For primes p such that p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), there is a lowest square m^2 such that m^2*p can be written in form x^2+n*y^2, where x and y are nonnegative integers (see A232529 and A232530).
If n is a prime congruent to 3 (mod 4), then a(n) = A232551(n).
The product of two numbers (prime or composite, same or different) which can be represented by the same quadratic form of discriminant = -4n can be written in the form x^2+n*y^2, as the following identity shows.
(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*Z+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(a*d-b*c)^2.
(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(b*d*(Y/X)+a*d+b*c)^2.
Note that for the latter equation, (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2)) and (b*d*(Y/X)+a*d+b*c) need not always be integers. If they are both integers, then it will be a second representation of the product of (X*a^2+Y*a*b+Z*b^2) and (X*c^2+Y*c*d+Z*d^2) in the form x^2+((X*Z)-(Y^2/4))*y^2.

Examples

			If n is a convenient number (A000926), then the only such available quadratic form is x^2+n*y^2.
For n = 11, every prime that is congruent to {0, 1, 3, 4, 5, 9} mod 11 can be represented by either of the two distinct primitive quadratic forms of discriminant = -44: x^2+11*y^2 or 3*x^2+2*x*y+4*y^2.
For n = 14, every prime that is congruent to {1, 2, 7, 9, 15, 23, 25, 39} mod 56 can be represented by either of the two distinct primitive quadratic forms of discriminant = -56: x^2+14*y^2 or 2*x^2+7*y^2.
For n = 17, every prime that is congruent to {1, 2, 9, 13, 17, 21, 25, 33, 49, 53} mod 68 can be represented by either of the two distinct primitive quadratic forms of discriminant = -68: x^2+17*y^2 or 2*x^2+2*x*y+9*y^2.
For n = 19, every prime that is congruent to {0, 1, 4, 5, 6, 7, 9, 11, 16, 17} mod 19 can be represented by either of the two distinct primitive quadratic forms of discriminant = -76: x^2+19*y^2 or 4*x^2+2*x*y+5*y^2.
For n = 20, every prime that is congruent to {1, 5, 9} mod 20 can be represented by either of the two distinct primitive quadratic forms of discriminant = -80: x^2+20*y^2 or 4*x^2+5*y^2.

Links

Crossrefs

Cf. A000003, A000926, A232529, A232530, A232551 (Number of distinct primitive quadratic forms of discriminant = -4*n needed to generate all primes p for which -n is a quadratic residue (mod p)).

A232551 Number of distinct primitive quadratic forms of discriminant -4n that exist such that every prime p for which -n is a quadratic residue mod p can be represented by one of them.

Original entry on oeis.org

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

Views

Author

V. Raman, Nov 26 2013

Keywords

Comments

This is closely related to the class number problem.
A quadratic form is primitive if the GCD of the coefficients is 1. For example, the quadratic form 2*x^2+4*y^2 is not primitive.
Two quadratic forms f(x,y) = a*x^2+b*x*y+c*y^2 and g(x,y) = p*x^2+q*x*y+r*y^2 are distinct (or inequivalent) if and only if one cannot be obtained by a linear transformation (of the variables x, y) from the other. For example, the three quadratic forms u(x,y) = 3*x^2+2*x*y+3*y^2, v(x,y) = 3*x^2+4*x*y+4*y^2 and w(x,y) = 3*x^2+10*x*y+11*y^2 are equivalent because v(x,y) = u(x+y,-y) and w(x,y) = v(x+y,y). Also, w(x,y) = u(x+2*y,-y). Similarly, the two quadratic forms s(x,y) = 8*x^2+9*y^2 and t(x,y) = 17*x^2+50*x*y+41*y^2 are equivalent because t(x,y) = s(x+2*y,x+y).
The quadratic form x^2+n*y^2 is one such form and the only such form if n = 1, 2, 3, 4, 7.
a(n) = 1 if and only if n = 1, 2, 3, 4, 7.
If n is a squarefree convenient number (A000926), a(n) represents the class number of the ring Z[sqrt(n)] if n == 1 (mod 4) or if n == 2 (mod 4) and the class number of the ring Z[(1+sqrt(n))/2] if n == 3 (mod 4) and this class number is a power of 2.
Any prime p such that -n is a quadratic residue mod p can be represented by exactly one of the a(n) distinct primitive quadratic forms of discriminant = -4n in at most four different ways (if n >= 2) or in at most eight different ways (if n = 1).
If n is a prime congruent to 3 (mod 4), then a(n) = A232550(n).
If p is a prime, p^2 does not divide n, and p > 2 if n == 3 (mod 8), then there is a multiple of p in which p is raised to an odd power which can be written in the form x^2+n*y^2 if and only if -n is a quadratic residue mod p.
The product of two numbers (prime or composite, same or different) which can be represented by the same quadratic form of discriminant = -4n can be written in the form x^2+n*y^2, as the following identity shows:
(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*Z+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(a*d-b*c)^2.
(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(b*d*(Y/X)+a*d+b*c)^2.
Note that for the latter equation, (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2)) and (b*d*(Y/X)+a*d+b*c) need not always be integers. If they are both integers, then it will be a second representation of the product of (X*a^2+Y*a*b+Z*b^2) and (X*c^2+Y*c*d+Z*d^2) in the form x^2+((X*Z)-(Y^2/4))*y^2.
This sequence is the same as taking every fourth number in A107628. - T. D. Noe, Jan 02 2014

Examples

			If n = 1, 2, 3, 4 or 7, then the only such available quadratic form is x^2+n*y^2.
For n = 5, every prime that is congruent to {1, 2, 3, 5, 7, 9} mod 20 can be represented by either of the two distinct primitive quadratic forms of discriminant = -20: x^2+5*y^2 or 2*x^2+2*x*y+3*y^2.
For n = 6, every prime that is congruent to {1, 2, 3, 5, 7, 11} mod 24 can be represented by either of the two distinct primitive quadratic forms of discriminant = -24: x^2+6*y^2 or 2*x^2+3*y^2.
For n = 10, every prime that is congruent to {1, 2, 5, 7, 9, 11, 13, 19, 23, 37} mod 40 can be represented by either of the two distinct primitive quadratic forms of discriminant = -40: x^2+10*y^2 or 2*x^2+5*y^2.

Links

Crossrefs

Cf. A000003, A000926, A232529, A232530, A232550 (Number of distinct primitive quadratic forms of discriminant = -4*n needed to generate all primes p for which p is a quadratic residue (mod 4*n) or p-n is a quadratic residue (mod 4*n)).

A232529 Least positive integer m such that for all primes p where p and p-n are quadratic residues (mod 4*n), (m^2)*p can be written as x^2+n*y^2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 2, 1, 1, 3, 1, 1, 3, 2, 1, 3, 1, 4, 2, 1, 5, 2, 2, 1, 3, 4, 1, 9, 1, 2, 3, 1, 5, 8, 1, 5, 3, 2, 2, 3, 3, 4, 3, 1, 1, 6, 1, 5, 9, 3, 2, 3, 5, 2, 6, 5, 1, 12, 1, 7, 9, 2, 4, 3, 1, 8, 3, 3, 7, 6, 2, 1, 9, 4, 1, 15, 3, 2, 3, 1, 25, 8, 2, 7, 7, 2, 2, 15
Offset: 1

Views

Author

V. Raman, Nov 25 2013

Keywords

Comments

If n is a convenient number (A000926), then a(n) = 1.
m is also the lowest nonzero integer such that m^2 can be generated by using all the inequivalent primitive quadratic forms of discriminant = -4n.

Examples

			For n = 59, all primes p such that p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n) are either of the form x^2+59*y^2 or 4*x^2+2*x*y+15*y^2 or 3*x^2+2*x*y+20*y^2 or 5*x^2+2*x*y+12*y^2 or 7*x^2+4*x*y+9*y^2.
We have (6^2)*(x^2+59*y^2) = (6*x)^2+59*(6*y)^2,
(6^2)*(4*x^2+2*x*y+15*y^2) = (12*x+3*y)^2 + 59*(3*y)^2,
(6^2)*(7*x^2+4*x*y+9*y^2) = (4*x+18*y)^2 + 59*(2*x)^2,
(6^2)*(3*x^2+2*x*y+20*y^2) = (7*x+22*y)^2 + 59*(x-2*y)^2,
(6^2)*(5*x^2+2*x*y+12*y^2) = (11*x+14*y)^2 + 59*(x-2*y)^2.
So, m = 6 satisfies this condition for n = 59: for all primes p such that p is a quadratic residue (mod 4*n) and p-n is a quadratic residue (mod 4*n), (m^2)*p can be written as x^2+n*y^2.
And m = 6 is the smallest value of m to satisfy this condition. So, a(59) = 6.

Crossrefs

Cf. A232530, A000926.

Formula

a(n)=sqrt(A232530(n)).
Showing 1-3 of 3 results.

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

Content is available under The OEIS End-User License Agreement.