The Hamilton-waterloo Problem: The Case Of 3- Cycles And 6-cycles

The Hamilton-Waterloo problem is to determine the existence of a 2-factorization of K(nn is odd)or K n? I (n is even,and I is a 1-factor),in which r of the 2-factors are isomorphic to a given 2-factor Q and s of the 2-factors are isomorphic to a given 2-factor R, r ,s∈N. If components of Q are cycles of length c and that of R are cycles of length d,then the corresponding instance of Hamilton-Waterloo problem is denoted by the existence of HW ( n; r , s; c,d ) or HW ( r , s; c,d ). The necessary conditions of the existence of HW ( r , s; c,d ) are:(1) If r > 0,then c n;If s > 0,then d n;(2) If n is odd, then r + s = n2?1;(3) If n is even, then r + s= n2? 1; Let ( ) {0,1,..., 1}2I n = n?,n is odd; I ( n ) = {0,1,..., n2? 1},n is even; obviously r , s∈I ( n).In this paper we consider the case when Q is a 3-cycle factor and R is a 6-cycle factor, which is the existence of HW ( n; r , s; 3,6). Obviously, The necessary conditions of the existence of HW ( n; r , s; 3,6) is 6 n,therefore let n = 6k,k∈N,then r + s = 3k ? 1, I (6 k ) = {0,1,...,3k ? 1}.Let ( )HW *(6 k ) = {s such t hat H W 6 k ; r , s; 3,6 exist} Obviously, HW *(6 k )? (I6k). This paper gives following results: When k≡0(mod6), {0;1,3, k ?1 ; k , k + 1, 2k ?1 ;2k ,2k + 2, 3k ? 2;3k ?1 } ?HW *(6k ) When k≡3(mod6), {0;1,3, 3k ? 2;3k ?1 } ? HW *(6k )Additionally, we give some results about the existence of HW (16 k + 4; r , s ; h ,4)in the end of this paper: {2,4, 4 k } ? HW *(16 k+ 4) Where h denote Hamilton-cycle consisted of cycles of length n, ( )HW *(16 k + 4) = {s such t hat H W 16 k + 4; r , s; h, 4 exist}...