Re: [理工] [離散]-遞迴
※ 引述《assassin88 (2010)》之銘言:
: For n >= 1, let an be the number of ways to write n as an ordered sum of
: positive integer where each summand is at least 2.
: 請問這一題要怎麼想?
: 完全沒有idea..麻煩指導了~感謝!
---
假設 x^n 的係數為題目所求
則考慮以下生成函數:
∞
f(x) = Σ (x^2 + x^3 + x^4 + ...)^k
k=1
∞ x^2 k
= Σ ( ───) if |x|<1
k=1 1 - x
∞ 2k ∞ m
= Σ x Σ C(-k,m)*(-x)
k=1 m=0
∞ ∞ m 2k+m
= Σ Σ C(-k,m)*(-1) x
k=1 m=0
(-k)(-k-1)...(-k-m+1)
where C(-k,m)≡ ──────────
m!
m
因此 x^n 的係數 = Σ C(-k,m)(-1)
n=2k+m
[n/2] n-2k
= Σ C(-k,n-2k)(-1)
k=1
即為所求
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例如當 n=8
有以下拆法:
8
6 + 2
2 + 6
5 + 3
3 + 5
4 + 4
4 + 2 + 2
2 + 4 + 2
2 + 2 + 4
3 + 3 + 2
3 + 2 + 3
2 + 3 + 3
2 + 2 + 2 + 2
共 13種
根據前面的推導
有 C(-1,6) + C(-2,4) + C(-3,2) + C(-4,0)
= 1 + 5 + 6 + 1
= 13 種拆法
希望沒搞錯題意 OTZ
--
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※ 編輯: doom8199 來自: 61.64.93.41 (02/07 20:15)
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