This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A341422 #21 Dec 22 2024 16:38:58 %S A341422 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 A341422 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 A341422 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 A341422 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 A341422 This gives the row lengths of the irregular triangle A343232. %C A341422 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 A341422 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 A341422 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 A341422 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 A341422 Cf. A000086 (with zeros), A002061, A034017, A343232. %K A341422 nonn,easy %O A341422 1,3 %A A341422 _Wolfdieter Lang_, Apr 08 2021