F(N) = SUM_(n1 + n2 + ... nk=N, ni >=3 ) {[C(N, n1)(n1-)!C(N, n2)(n2-1)!...C(N, nk)(nk-1)!)]/D(k)]},
where D(k) is the duplication factor for the decomposition of N = n1 + ... + nk, nj >=3. That is, suppose there are m different numbers in the set {n1, n2, ..., nk}, and they occur in the set l1, l2, ... lm times, (l1 + ... + lm = k), then
D(k) = (l1)!*(l2)!*(l3)!*...(lm)!.
thanks. So the general solution should be
所有跟帖:
• F(N) = SUM_(n1 + n2 + ... nk=N, ni >=3 ) {[C(N, n1)C(N-n1,n2).. -jinjing- ♀ (66 bytes) () 05/06/2012 postreply 09:37:01
• ... {[C(N, n1)C(N-n1,n2)..C(N-n1-n2-...n(k-1,nk)/D(n1,n2m,,,nk)] -jinjing- ♀ (0 bytes) () 05/06/2012 postreply 09:41:59