%I #18 Aug 11 2015 01:31:23
%S 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,
%T 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,
%U 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
%N 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.
%C If n is a convenient number (A000926), then a(n) = 1.
%C 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.
%F a(n)=sqrt(A232530(n)).
%e 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.
%e We have (6^2)*(x^2+59*y^2) = (6*x)^2+59*(6*y)^2,
%e (6^2)*(4*x^2+2*x*y+15*y^2) = (12*x+3*y)^2 + 59*(3*y)^2,
%e (6^2)*(7*x^2+4*x*y+9*y^2) = (4*x+18*y)^2 + 59*(2*x)^2,
%e (6^2)*(3*x^2+2*x*y+20*y^2) = (7*x+22*y)^2 + 59*(x-2*y)^2,
%e (6^2)*(5*x^2+2*x*y+12*y^2) = (11*x+14*y)^2 + 59*(x-2*y)^2.
%e 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.
%e And m = 6 is the smallest value of m to satisfy this condition. So, a(59) = 6.
%Y Cf. A232530, A000926.
%K nonn,uned
%O 1,11
%A _V. Raman_, Nov 25 2013
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