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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

An upper bound on the number of solutions appears to be 1.5*sqrt(n). - T. D. Noe, Jun 14 2006

a(n) is also the total number of distinct quadratic forms of discriminant -4n. A232551 counts only the primitive quadratic forms of discriminant -4n (those with all coefficients pairwise coprime) and A234287 includes those by which some prime can be represented (those with all coefficients pairwise coprime or coefficient of x^2 is prime or coefficient of y^2 is prime). This sequence includes all quadratic forms like 2x^2 + 2xy + 4y^2 and 2x^2 + 4y^2 which are non-primitive and those like 4x^2 + 2xy + 4y^2 and 4x^2 + 4xy + 4y^2 by which no prime can be represented (those with no restrictions). - V. Raman, Dec 24 2013

This is similar to A232551, except that this includes non-primitive quadratic forms like 2x^2+2xy+4y^2 and 2x^2+4y^2 because the prime 2 can be represented by both of them. But unlike A067752, we do not include quadratic forms like 4x^2+2xy+4y^2 and 4x^2+4xy+4y^2 by which no prime can be represented.

So, when n == 3 (mod 4), this includes the additional non-primitive quadratic form 2x^2+2xy+((n+1)/2)y^2 and when p^2 divides n, where p is prime, this includes the additional non-primitive quadratic form px^2+(n/p)y^2.

If p is a prime and if p^2 does not divide n, then there exist a unique non-primitive quadratic form of discriminant = -4n by which p can be represented if and only if -n is a quadratic residue (mod p) and there exists a multiple of p which can be written in the form x^2+ny^2 in which p appears raised to an odd power, except when p = 2 and n == 3 (mod 8).

If n is a convenient number (A000926), the set of residue classes (mod 4n) that a prime p represented by x^2+n*y^2 belong to are those for which 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 non-convenient numbers n, some of the primes in these set of residue classes (mod 4n) can be represented by x^2+n*y^2, but not all.

A prime p such that p^2 does not divide n, can be represented by some primitive quadratic form of discriminant = -4n, if and only if -n is a quadratic residue (mod p).

A prime p can be represented by some quadratic form of discriminant = -4n if and only if there is a multiple of p that can be written in the x^2+n*y^2 form, in which prime factor of p appears raised to an odd power or if p = 2 and n == 3 (mod 4).

a(n) is always a divisor of 4n.

If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 4n.

If p^2 divides n for some prime p, a(n) is a divisor of (4n)/p.

If n == 3 (mod 8), then a(n) is a divisor of n because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.

If n == 7 (mod 8), then a(n) is a divisor of 2n because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.

Please look into A234001 for a more detailed description.

If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 1.

If p^2 divides n for some prime p, a(n) is a multiple of p.

If n == 3 (mod 8), then a(n) is a multiple of 4 because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.

If n == 7 (mod 8), then a(n) is a multiple of 2 because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.