OFFSET
1,3
COMMENTS
For A358946(1) = 1 one uses m(1) = 0.
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].
For details on reduced, primitive or parallel forms, proper representations and proper equivalence see A358946 with the linked papers where references are given.
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).
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.
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.
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.
REFERENCES
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, Theorems 5.28, pp. 118-119, and 5.30, pp. 121-122.
EXAMPLE
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.
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].
The four fundamental representations of F_p = [1, 4, -3] for k = 57 are (11, 16), (5, 4), (6, 7) and (6, 1), respectively.
CROSSREFS
KEYWORD
nonn
AUTHOR
Wolfdieter Lang, Jan 10 2023
STATUS
approved