首先
h
n
=
∑
i
h
i
h
n
−
i
−
1
h_n=\sum_{i}h_ih_{n-i-1}
hn=i∑hihn−i−1
写出
h
h
h 的母函数
H
(
x
)
H(x)
H(x)
那么
H
(
x
)
=
H
2
(
x
)
x
+
1
,
H
(
x
)
=
1
−
1
−
4
x
2
x
H(x)=H^2(x)x+1,H(x)=\frac{1-\sqrt{1-4x}}{2x}
H(x)=H2(x)x+1,H(x)=2x1−1−4x
(解二元一次方程取符号时候要看是否收敛)
引入牛顿二项式
(
x
+
y
)
α
=
∑
k
=
0
∞
(
α
k
)
x
α
−
k
y
k
(x+y)^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{\alpha-k}y^{k}
(x+y)α=k=0∑∞(kα)xα−kyk
其中
(
α
k
)
=
∏
i
=
1
k
α
−
i
+
1
i
\binom{\alpha}{k}=\prod_{i=1}^{k}\frac{\alpha - i + 1}{i}
(kα)=i=1∏kiα−i+1
展开可以得到
H
(
x
)
=
1
−
∑
k
=
0
∞
(
1
2
k
)
(
−
4
x
)
k
2
x
H(x)=\frac{1-\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(-4x)^k}{2x}
H(x)=2x1−∑k=0∞(k21)(−4x)k
=
−
1
2
∑
k
=
0
∞
(
1
2
k
+
1
)
(
−
4
)
k
+
1
x
k
=-\frac{1}{2}\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k+1}(-4)^{k+1}x^k
=−21k=0∑∞(k+121)(−4)k+1xk
=
2
∑
k
=
0
∞
(
1
2
k
+
1
)
(
−
4
x
)
k
=2\sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k+1}(-4x)^k
=2k=0∑∞(k+121)(−4x)k
那么
h
n
=
2
(
1
2
n
+
1
)
(
−
4
x
)
n
=
2
∏
i
=
0
n
(
1
2
−
i
)
(
n
+
1
)
!
(
−
1
)
n
2
2
n
h_n=2\binom{\frac{1}{2}}{n+1}(-4x)^n=2\frac{\prod_{i=0}^{n}(\frac{1}{2}-i)}{(n+1)!}(-1)^n2^{2n}
hn=2(n+121)(−4x)n=2(n+1)!∏i=0n(21−i)(−1)n22n
=
∏
i
=
0
n
(
1
−
2
i
)
(
n
+
1
)
!
(
−
1
)
n
2
n
=
∏
i
=
1
n
(
2
i
−
1
)
(
n
+
1
)
!
2
n
=
(
2
n
−
1
)
!
!
(
n
+
1
)
!
2
n
=\frac{\prod_{i=0}^{n}(1-2i)}{(n+1)!}(-1)^n2^n=\frac{\prod_{i=1}^{n}(2i-1)}{(n+1)!}2^n=\frac{(2n-1)!!}{(n+1)!}2^n
=(n+1)!∏i=0n(1−2i)(−1)n2n=(n+1)!∏i=1n(2i−1)2n=(n+1)!(2n−1)!!2n
而
(
2
n
−
1
)
!
!
+
2
n
n
!
=
(
2
n
)
!
(2n-1)!!+2^nn!=(2n)!
(2n−1)!!+2nn!=(2n)!
所以
h
n
=
(
2
n
)
!
n
!
(
n
+
1
)
!
=
1
n
+
1
(
2
n
n
)
h_n=\frac{(2n)!}{n!(n+1)!}=\frac{1}{n+1}\binom{2n}{n}
hn=n!(n+1)!(2n)!=n+11(n2n)
完美解决
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