The number of triangles in 2-factorizations of K2n minus a 1-factor

A 2-factorization of K-n is a partition of the edge set of K-n into 2-factors. Given an arbitrary 2-factorization F = {F-1, F-2,..., Fn-1} of K-2n, let delta(i) be the number of triangles contained in F-i and let delta = Sigma delta(i). Then.F is said to be a 2-factorization with delta triangles. Denote by Delta(2n), the set of all delta such that there exists a 2-factorization with delta triangles. Let P-Delta(2n) = {{0, 1,..., M-2n} if 2n equivalent to 2 or 4 (mod 6), {0, 1,...,M-2n}\{M2n - 1} if 2n equivalent to 0 (mod 6), where M-2n = {(n-1)(2n-5)/3 if 2nd equivalent to 2 (mod 6), 2(n-1)(n-2)/3 if 2nd equivalent to 4 (mod 6), 2n(n-1)/3 if 2nd equivalent to 0 (mod 6), In this paper, we consider the problem of constructing 2-factorization of K-2n containing a specified number of triangles. We proved that apart from some exceptions Delta(2n) = P-Delta(2n).