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A225229 Numbers n such that if some integer N can be written in the form (a/2)^2+n*(b/2)^2 for integers a and b, then every prime factor P of N which occurs to an odd power can also be written in the form (c/2)^2+n*(d/2)^2 for integers c and d. 0
1, 2, 3, 4, 7, 8, 11, 12, 19, 43, 67, 163 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Alternately, values of n such that if for some number N, 4*N can be written as x^2+n*y^2, then for every prime factor p of N which occurs to an odd power, 4*p can be written as x^2+n*y^2.

For these values of n, for all primes p for which -n is a quadratic residue (mod p), 4*p can be written as x^2+n*y^2.

For n = 1, 2, 3 all primes p for which -n is a quadratic residue (mod p) can be written as x^2+n*y^2, with 4*p = (2*x)^2+n*(2*y)^2.

n = 1, 2, 3 are also the only values of n for which if some positive integer N can be written in the form a^2+n*b^2, for some integers a and b, then every prime factor P of N which occurs to an odd power can be written in the form c^2+n*d^2, for some integers c and d.

For n = 4, 7 all primes p for which -n is a quadratic residue (mod p) except for p = 2 can be written as x^2+n*y^2, with 4*p = (2*x)^2+n*(2*y)^2. For p = 2, 4 * 2 = 8 = 2^2 + 4*1^2 = 1^2 + 7*1^2.

For n = 8, all primes p (except for p = 2), for which -n is a quadratic residue (mod p) are generated by either x^2+8*y^2 or 3*x^2+2*x*y+3*y^2. We have 4*(x^2+8*y^2) = (2*x)^2+8*(2*y)^2 and 4*(3*x^2+2*x*y+3*y^2) = (2*x-2*y)^2+8*(x+y)^2. For p = 2, 4 * 2 = 8 = 0^2 + 8*1^2.

For n = 12, all primes p (except for p = 2), for which -n is a quadratic residue (mod p) are generated by either x^2+12*y^2 or 3*x^2+4*y^2. We have 4*(x^2+12*y^2) = (2*x)^2+12*(2*y)^2 and 4*(3*x^2+4*y^2) = (4*y)^2+12*x^2. We do not need to consider the prime p = 2 because numbers of form x^2+12*y^2 cannot contain prime factor of 2 raised to an odd power, as 12 is of the form 4^s*(8*t+3).

For n = 11, all primes p for which -n is a quadratic residue (mod p) are generated by either x^2+11*y^2 or 3*x^2+2*x*y+4*y^2. We have 4*(x^2+11*y^2) = (2*x)^2+11*(2*y)^2 and 4*(3*x^2+2*x*y+4*y^2) = (x+4*y)^2+11*x^2.

For n = 19, all primes p for which -n is a quadratic residue (mod p) are generated by either x^2+19*y^2 or 4*x^2+2*x*y+5*y^2. We have 4*(x^2+19*y^2) = (2*x)^2+19*(2*y)^2 and 4*(4*x^2+2*x*y+5*y^2) = (4*x+y)^2+19*y^2.

For n = 43, all primes p for which -n is a quadratic residue (mod p) are generated by either x^2+43*y^2 or 4*x^2+2*x*y+11*y^2. We have 4*(x^2+43*y^2) = (2*x)^2+43*(2*y)^2 and 4*(4*x^2+2*x*y+11*y^2) = (4*x+y)^2+43*y^2.

For n = 67, all primes p for which -n is a quadratic residue (mod p) are generated by either x^2+67*y^2 or 4*x^2+2*x*y+17*y^2. We have 4*(x^2+67*y^2) = (2*x)^2+67*(2*y)^2 and 4*(4*x^2+2*x*y+17*y^2) = (4*x+y)^2+67*y^2.

For n = 163, all primes p for which -n is a quadratic residue (mod p) are generated by either x^2+163*y^2 or 4*x^2+2*x*y+41*y^2. We have 4*(x^2+163*y^2) = (2*x)^2+163*(2*y)^2 and 4*(4*x^2+2*x*y+41*y^2) = (4*x+y)^2+163*y^2.

LINKS

Table of n, a(n) for n=1..12.

EXAMPLE

n = 27 is not a member of this sequence because N = 27 = (0/2)^2 + 27*(2/2)^2 is of the form (x/2)^2+27*(y/2)^2, but for the prime factor 3 which appears to an odd power in N = 27, 3 is not of the form (x/2)^2+27*(y/2)^2 because 4 * 3 = 12 is not of the form x^2+27*y^2 (also 3 itself is not of the form x^2+27*y^2).

CROSSREFS

Cf. A133675, A003173 (squarefree values of this sequence).

Sequence in context: A271441 A328421 A003508 * A239389 A256219 A078662

Adjacent sequences:  A225226 A225227 A225228 * A225230 A225231 A225232

KEYWORD

nonn,fini,full

AUTHOR

V. Raman, Apr 30 2013

STATUS

approved

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Last modified January 24 00:05 EST 2022. Contains 350515 sequences. (Running on oeis4.)