thanks. So the general solution should be

来源: wxcfan123 2012-05-06 08:28:32 [] [旧帖] [给我悄悄话] 本文已被阅读: 0 次 (416 bytes)
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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)!.

所有跟帖: 

F(N) = SUM_(n1 + n2 + ... nk=N, ni >=3 ) {[C(N, n1)C(N-n1,n2).. -jinjing- 给 jinjing 发送悄悄话 (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- 给 jinjing 发送悄悄话 (0 bytes) () 05/06/2012 postreply 09:41:59

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