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%I #36 Oct 17 2012 08:22:32
%S 0,1,1,1,3,2,2,3,3,4,3,4,6,3,3,4,7,5,5,7,6,5,4,6,8,6,6,8,9,5,6,8,9,8,
%T 5,9,11,6,6,11,12,6,8,9,12,7,6,10,11,9,8,15,12,10,8,10,13,10,8,10,15,
%U 7,9,12,16,9,10,15,12,10,8,15,18,10,9,16,12,8,11,15,17,12,9,16,19,10,9,16,18,13,12,13,14,11,9,15,21,10,14,20
%N Total number of solutions to the equation x^2+k*y^2 = n with x > 0, y > 0, k > 0. (Order matters for the equation x^2+y^2 = n).
%C If the equation x^2+y^2 = n has got two solutions (x, y), (y, x) then they will be counted differently.
%C No solutions can exist for the values of k >= n.
%C a(n) is the same as A216672(n) when n is not the sum of two positive squares.
%C But when n is the sum of two positive squares, the ordered pairs for the equation x^2+y^2 = n count.
%C For example,
%C 10 = 3^2 + 1^2.
%C 10 = 1^2 + 3^2.
%C 10 = 2^2 + 6*1^2.
%C 10 = 1^2 + 9*1^2.
%C So a(10) = 4. On the other hand, for the sequence A216672, the ordered pair 3^2+1^2 and 1^2+3^2 will be counted as the same, and so A216672(10) = 3.
%o (PARI) for(n=1, 100, sol=0; for(k=1, n, for(x=1, n, if((issquare(n-k*x*x)&&n-k*x*x>0), sol++))); print1(sol", ")) /* _V. Raman_, Oct 16 2012 */
%Y Cf. A216503, A216504, A216505.
%Y Cf. A216672 (a variant of this sequence, when the order does not matter for the equation x^2+y^2 = n, i.e. if the equation x^2+y^2 = n has got two solutions (x, y), (y, x) then they will be counted as the same).
%Y Cf. A216672, A216673, A216674, A217840, A217956.
%K nonn
%O 1,5
%A _V. Raman_, Oct 16 2012
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