谢,K>=3,OK请看.m=2,不用做.原式可能有小笔误.

M=2,OK,设K时OK, N(k)>=c(k,2)(k/2)^2/c(k,2), N(K+1)=[c(k+1,1)c(k,2)*(k/2)^2]/c((k+1),2)=...>=[(k+1)^2/2]^2. 证OK

M=K+1时时,有[a1,(a2,.....ak(+1))],[a2,(a1,......,ak(+1))],......,[a(k+1),(a1,...ak)],k+1种.c(k+1)

We have N(K+1)=[c(k+1,1)c(k,2)*(k/2)^2]/c((k+1),2)=[(k+1)*(k(k-1))/2]*[k*k/4]/[(k+1)*k/2]=k*k(k-1)/4

k>=3,We have >=[(k+1)^2/2]^2.

所有跟帖： 

• should prove case m=3 as the first step of mathematics induction -wxcfan123- ♂ (0 bytes) () 02/06/2015 postreply 19:56:54

• Thanks,though it is trivial. -jinjing- ♀ (0 bytes) () 02/07/2015 postreply 08:08:36

• 望不吝赐教。之所以好奇，是因为整个证明与LCM无关。太神了。 -wxcfan123- ♂ (0 bytes) () 02/07/2015 postreply 10:38:56