Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #16 Dec 16 2013 12:12:51
%S 1,1,1,1,1,1,1,1,1,1,4,1,1,9,1,1,9,1,4,4,1,1,9,1,1,9,4,1,9,1,16,4,1,
%T 25,4,4,1,9,16,1,81,1,4,9,1,25,64,1,25,9,4,4,9,9,16,9,1,1,36,1,25,81,
%U 9,4,9,25,4,36,25,1,144,1,49,81,4,16,9,1,64,9,9,49,36,4,1,81,16,1,225,9,4,9,1,625,64,4,49
%N Least square m^2 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.
%C If n is a convenient number (A000926) then a(n) = 1.
%C m^2 is also the smallest positive square that can be generated by using all the inequivalent primitive quadratic forms of discriminant = -4n.
%F a(n)=A232529(n)^2.
%e For n = 11, 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+11*y^2 or 3*x^2+2*x*y+4*y^2.
%e 4*(x^2+11*y^2) = (2*x)^2+11*(2*y)^2, 4*(3*x^2+2*x*y+4*y^2) = (x+4*y)^2+11*x^2. Also, 4 is the smallest square to satisfy this condition. So, a(11) = 4.
%e For n = 14, 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+14*y^2 or 2*x^2+7*y^2.
%e 9*(x^2+14*y^2) = (3*x)^2+14*(3*y)^2, 9*(2*x^2+7*y^2) = (2*x+7*y)^2+14*(x-y)^2 = (2*x-7*y)^2+14*(x+y)^2. Also, 9 is the smallest square to satisfy this condition. So, a(14) = 9.
%e For n = 17, 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+17*y^2 or 2*x^2+2*x*y+9*y^2.
%e 9*(x^2+17*y^2) = (3*x)^2+17*(3*y)^2, 9*(2*x^2+2*x*y+9*y^2) = (x+9*y)^2+17*x^2. Also, 9 is the smallest square to satisfy this condition. So, a(17) = 9.
%e For n = 19, 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+19*y^2 or 4*x^2+2*x*y+5*y^2.
%e 4*(x^2+19*y^2) = (2*x)^2+19*(2*y)^2, 4*(4*x^2+2*x*y+5*y^2) = (4*x+y)^2+19*y^2. Also, 4 is the smallest square to satisfy this condition. So, a(19) = 4.
%e For n = 20, 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+20*y^2 or 4*x^2+5*y^2.
%e 4*(x^2+20*y^2) = (2*x)^2+20*(2*y)^2, 4*(4*x^2+5*y^2) = (4*x)^2+20*y^2. Also, 4 is the smallest square to satisfy this condition. So, a(20) = 4.
%Y Cf. A232529, A000926.
%K nonn,uned
%O 1,11
%A _V. Raman_, Nov 25 2013