%I #18 Feb 03 2023 01:32:30

%S 1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,2,2,2,2,2,2,2,4,2,2,2,2,2,2,2,2,2,

%T 2,2,2,4,2,2,2,2,2,4,4,2,4,2,2,2,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2,4,4,2,

%U 2,2,4,2,2,2,2,2,4,2,4,2,2,2,2,4,2,2,2,2,4,4,2,2,2,4,2,2

%N a(n) is the number of solutions of the congruence j^2 + j + 1 == 0 (mod k = A034017(n+1)), for j from {0, 1, 2, ..., k-1}, for n >= 1.

%C This gives the row lengths of the irregular triangle A343232.

%C This sequence gives the number of representative parallel primitive forms (rpapfs) of the positive definite binary quadratic form F = x^2 + x*y + y^2 (with Discriminant Disc = -3) representing positive integers k. Only certain odd k, namely k = k(n) = A034017(n+1), for n >= 1, have proper solutions F = k.

%C Positive definite binary quadratic primitive forms F = [a, b, c], with a > 0 and gcd(a, b, c) = 1, with odd discriminants Disc = b^2 - 4*a*c = -D < 0, that is, D == 3 (mod 4), and representation of positive integers k have representative parallel primitive forms (rpapfs) Fpa(D,k;j) = [k, 2*j+1, (j^2 + j + (D+1)/4)/k].

%C Each rpapf produces a trivial proper solution to F = k, obtained from the trivial solution of Fpa(D,k;j) = k by (x, y) = (1,0), via equivalence transformations of determinant +1 achieved by applying the inverse of products of matrices R(t) = Mat([0,-1], [1t]]) for certain values t. The R(t) transformations are used to obtain from a primitive form F = [a, b, c] the equivalent so-called unique half-reduced (right) neighbor form F' = [c, -b + 2*c*t, a - b*t + c*t^2], with the choice t = ceiling((b/c - 1)/2). (c > 0 because a > 0 for positive definite forms with D > 0.)

%F a(n) = |M(k(n)|, with the set M(k(n)) := {j from {0, 1 , ..., k(n)-1} | j^2 + j + 1 == 0 (mod k(n))}, where j^2 + j + 1 = 2*T(j) + 1 = A002061(j+1) and k(n) = A034017(n+1), for n >= 1.

%Y Cf. A000086 (with zeros), A002061, A034017, A343232.

%K nonn,easy

%O 1,3

%A _Wolfdieter Lang_, Apr 08 2021