%I #11 Jan 12 2023 01:53:08
%S 1,1,2,2,2,2,2,2,2,4,2,2,2,4,2,2,2,4,2,2,4,2,2,4,4,2,2,2,2,2,2,2,4,4,
%T 2,2,2,2,4,2,4,2,2,4,2,4,2,2,2,2,2,4,2,2,2,4,2,2,2,2,2,4,4,4,4,4,2,2,
%U 2,2,2,2,4,4,4,2,4,2,4,4,4,2,4,4,2,4,4
%N a(n) = 2^m(n), where m(n) is the number of distinct primes, neither 2 nor 7, dividing A358946(n).
%C For A358946(1) = 1 one uses m(1) = 0.
%C a(n) gives the number of representative parallel primitive forms (rpapfs) of Disc = 28 representing k = A358946(n), that is the number of proper fundamental representations X = (x, y) of each indefinite primitive binary quadratic form of discriminant Disc = 28 = 2^2*4 which is properly equivalent to the reduced principal form F_p = x^2 + 4*x*y - 3*y^2, denoted by F_p = [1, 4, -3].
%C For details on reduced, primitive or parallel forms, proper representations and proper equivalence see A358946 with the linked papers where references are given.
%C The proof uses the formula for finding the rpapfs of Disc = 28 representing k, namely c = c(j, k) =(j^2 - 7)/k, for j from {0, 1, ..., k-1}, with the form(s) FPa(k) = [k, 2*j, c(j,k)]. Thus, j^2 - 7 == 0 (mod k).
%C For the representation k = 1 = A358946(1) there is the unique j = 0 solution with FPa(1) = [1, 0, -7], the neither reduced nor half-reduced Pell form FPell of Disc = 28. FPell is properly equivalent to the reduced F_p.
%C For k = 2 = A358946(2) the unique solution is j = 1 (-1 == 1 (mod 2)) with FPa(2) = [2, 2,-3], one of the four reduced forms of the principal cycle Cy(28) including F_p as a form. There is no lifting of 2 to powers of 2 (see Apostol, Theorem 5.30, p. 121), hence no factor 2^q, for q >= 2 appears in A358946.
%C For powers of primes (not 2 or 7) in A358946(n) one has for each prime p two solutions j and p-j of j^2 - 7 == 0 (mod p), and each can be lifted uniquely to powers of this prime. Thus each power of a prime counts as 2, like for the prime. See some examples below.
%D Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, Theorems 5.28, pp. 118-119, and 5.30, pp. 121-122.
%e k = 57 = 3*19 = A358946(10): One has for k = 3 the two solutions j = 1, 2, giving the parallel forms FPa(3)_1 = [3, 2, -2], belonging to the cycle Cyhat(28), thus 3 is not in A358946 but 3 = A359476(1) for the representaion of k = -3, and FPa(3)_2 = [3, 4, -1], also in Cyhat(28). The lifting of the solutions from k = 3 to k = 3^2 = 9 is possible uniquely because 2*j = 2 and 2*j = 4 are both not congruent to 0 (mod 3). See the two parallel forms for k = 9 above. For k = 19 one has j = 8 and j = 11, with the forms FPa(19)_1 = [19, 16, 3] and FPa(19)_2 = [19, 22, 6], both reaching the cycle Cyhat(28) after R-transformations with t = 3 and first t = -2 then t = 3, respectively. Thus k = 19 belongs to A359476, not A358946.
%e The four solutions for k = 57 are j = 8, 11, 46, 49, with parallel forms [57, 16, 1], [57, 22, 2], [57, 92, 37] and [57, 98, 42].
%e The four fundamental representations of F_p = [1, 4, -3] for k = 57 are (11, 16), (5, 4), (6, 7) and (6, 1), respectively.
%Y Cf. A358946, A359476, A359477.
%K nonn
%O 1,3
%A _Wolfdieter Lang_, Jan 10 2023