作者PttFund (批踢踢基金只进不出)
看板Math
标题[离散] 离散(3)
时间Sun Jul 24 16:57:31 2005
The number f(n) of steps required to solve the "chinese rings
puzzle" with n ring satisfies f(1) = 1 and
f(n+1) = 2f(n) if n is odd,
2f(n) + 1 if n is even.
Prove that f(n+2) = f(n+1) + 2f(n) + 1. Hence or otherwise
find a formula for f(n) in term of n.
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请到(P)LAY-->(P)AY-->(0)GIVE-->PttFund-->吧
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◆ From: 140.112.218.142
1F：推 JGU:2^(n+1)/3 + (-1)^(n+1)/6 - 1/2 61.229.112.108 07/25
2F：→ eggsu :我也做出来了，谢谢JGU的答案让我能把答案更化简 04/29 23:10
3F：→ eggsu :本来我做出来的是 04/29 23:10
4F：→ eggsu :当 n 是奇数时，f(n) = 2^(n+1)/3 - 1/3 04/29 23:11
5F：→ eggsu :当 n 是偶数时，f(n) = 2^(n+1)/3 - 2/3 04/29 23:11
6F：→ eggsu :加强了对振荡数列的体会，感恩啊！ 04/29 23:13