M=2,OK,設K時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. 證OK
M=K+1時時,有[a1,(a2,.....ak(+1))],[a2,(a1,......,ak(+1))],......,[a(k+1),(a1,...ak)],k+1種.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.
謝,K>=3,OK請看.m=2,不用做.原式可能有小筆誤.
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
• 望不吝賜教。之所以好奇,是因為整個證明與LCM無關。太神了。 -wxcfan123- ♂ (0 bytes) () 02/07/2015 postreply 10:38:56