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.

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

1,11

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.

Links

Table of n, a(n) for n=1..101.

Formula

a(n)=sqrt(A232530(n)).

Example

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.

Sequence in context: A117502 A212630 A030360 * A095374 A352063 A300650

Adjacent sequences: A232526 A232527 A232528 * A232530 A232531 A232532

Keyword

nonn,uned

Author

V. Raman, Nov 25 2013

Status

approved