1 Answer

Alphager · Answer 1 · 2020-12-25T00:51:23+0000

This sequence has generating function

$F(x)=\displaystyle\sum_(k\ge0)k^3x^k$

(if we include
k=0 for a moment)

Recall that for
|x|<1 , we have

$\displaystyle\frac1{1-x}=\sum_(k\ge0)x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_(k\ge0)kx^(k-1)=\frac1x\sum_(k\ge0)kx^k$

$\implies\frac x{(1-x)^2}=\displaystyle\sum_(k\ge0)kx^k$

Take the derivative again:

$\displaystyle((1-x)^2+2x(1-x))/((1-x)^4)=\sum_(k\ge0)k^2x^(k-1)=\frac1x\sum_(k\ge0)k^2x^k$

$\implies\displaystyle(x+x^2)/((1-x)^3)=\sum_(k\ge0)k^2x^k$

Take the derivative one more time:

$\displaystyle((1+2x)(1-x)^3+3(x+x^2)(1-x)^2)/((1-x)^6)=\sum_(k\ge0)k^3x^(k-1)=\frac1x\sum_(k\ge0)k^3x^k$

$\implies\displaystyle(x+4x^3+x^3)/((1-x)^4)=\sum_(k\ge0)k^3x^k$

so we have

$\boxed{F(x)=(x+4x^3+x^3)/((1-x)^4)}$

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