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%I #11 Mar 30 2021 16:45:13
%S 1,4,5,8,13,17,20,25,29,37,40,41,52,53,61,65,68,73,85,89,97,100,101,
%T 104,109,113,116,125,136,137,145,148,149,157,164,169,173,181,185,193,
%U 197,200,205,212,221,229,232,233,241,244,257,260,265,269,277,281,289,292,293,296
%N The positive integer numbers k represented properly by the binary quadratic form x^2 + 4*y^2.
%C If also improper solutions of the Diophantine equation X^2 + Y^2 = k, with positive integer number k are taken into account one can obtain the present solutions provided X or Y are even. E.g., k = 4 has only improper solutions like (X, Y) = (0, pm2) or (pm2, 0) (pm stands for +1 or -1). So 4 is not a member of A008784, but in the present sequence it appears from (x, y) = (0, pm1) obtained from the first (X, Y) solution by y = Y/2.
%C The number k = 2 = A008784(2) is not represented here because there is only the proper solution (X, Y) = (pm1, pm1).
%C The number of solutions m(k = a(n)), up to an overall sign change in x and y, is given by m(1) = 1, m(4) = 1, m(8) = 2 and for k = 4^a*8^b*Product_{j=1..P1} (p1_j)^e1_j, with (a,b) from {(0, 0), (1, 0), (0, 1)}}, primes p1_j congruent to 1 (mod 4) (from A002144) and nonnegative exponents e1_j, it is m(k) = 2^(b + P1).
%C The primitive parallel binary quadratic forms of discriminant -16 = -4*4 representing positive integer numbers k are obtained by solving the Diophantine equation j^2 + 4 == 0 (mod k), for j from {0, 1, ..., k-1}. This gives for k = 1, 2, 4, and 8 the solutions j = 0, 0, {0, 2}, and {2, 6}, respectively. No larger powers of 2 have solutions. No lifting is possible (see Apostol, Theorem 5.30). For odd primes k the Legendre symbol (-4, k) = +1 exactly for k = prime == 1 (mod 4) (from the Legendre symbol (-1, prime) = +1 only for these primes A008784).
%C These parallel forms are given by (k, 2*j(k), c(j(k))), with c(j(k)) = (j(k)^2 + 4)/k.
%C There is only one primitive reduced form for discriminant -16, namely the principal form (1, 0, 4) (see the Buell reference p. 20). Thus each parallel form is equivalent (with a determinant +1 transformation) to this principal form, and gives a proper solution.
%D Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp. 120-122.
%D D. A. Buell, Binary Quadratic Forms, Springer, 1989, p. 20.
%F a(n) = x(n)^2 + (2*y(n))^2, with gcd(x(n), y(n)) = 1, for n >= 1.
%e Proper solutions (x, y) (up to overall sign flip) for various k = a(n):
%e a(2) = 4: (1, 0), m(4) = 1 (a = 1, b = 0, P1 = 0), (2, 0) is not a proper solution);
%e a(4) = 8: (2, pm1): (pm stands for +1 or -1), m(8) = 2 (a = 0, b = 1, P1 = 0);
%e a(7) = 20 = 4*5: (4, pm1), m(20) = 2 (a = 1, b = 0, P1 = 1), (m(4) = 1);
%e a(8) = 25 = 5^2: (3, pm2), m(25) = 2 (a = 0, b = 0, P1 = 1);
%e a(42) = 200 = 8*5^2: (2, pm7), (14, pm1), m(200) = 4 (a = 0, b = 1, P1 = 1).
%Y Cf. Discriminants -4: A008784, -8 A057127, -12 A244819.
%K nonn
%O 1,2
%A _Wolfdieter Lang_, Mar 19 2021
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