Respuesta :

LammettHash
Let

[tex]H(x)=\displaystyle\sum_{n\ge0}h_nx^n[/tex]

be the generating function for the sequence [tex]h_n[/tex]. Then

[tex]h_n=3h_{n-1}-4n[/tex]
[tex]\displaystyle\sum_{n\ge1}h_nx^n=3\sum_{n\ge1}h_{n-1}x^n-4\sum_{n\ge1}nx^n[/tex]
[tex]\displaystyle\sum_{n\ge0}h_nx^n-h_0=3x\sum_{n\ge0}h_nx^n-4x\sum_{n\ge0}nx^{n-1}[/tex]
[tex]\displaystyle H(x)-h_0=3xH(x)-4x\frac{\mathrm d}{\mathrm dx}\sum_{n\ge0}x^n[/tex]
[tex](1-3x)H(x)=h_0-4x\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right][/tex]
[tex](1-3x)H(x)=h_0-\dfrac{4x}{(1-x)^2}[/tex]
[tex]H(x)=\dfrac{h_0}{1-3x}-\dfrac{4x}{(1-3x)(1-x)^2}[/tex]

Decompose the latter term into partial fractions:


[tex]-\dfrac{4x}{(1-3x)(1-x)^2}=\dfrac2{(1-x)^2}+\dfrac1{1-x}-\dfrac3{1-3x}[/tex]

so that

[tex]H(x)=\dfrac{h_0-3}{1-3x}+\dfrac1{1-x}+\dfrac2{(1-x)^2}[/tex]
[tex]\implies H(x)=\displaystyle\sum_{n\ge0}(h_0-3)3^nx^n+\sum_{n\ge0}x^n+\sum_{n\ge0}2nx^n[/tex]
[tex]\implies H(x)=\displaystyle\sum_{n\ge0}\bigg((h_0-3)3^n+2n+1\bigg)x^n[/tex]
[tex]\implies h_n=(h_0-3)3^n+2n+1[/tex]

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