If $\sqrt[3]{1 − 2x}$ is the generating function of a sequence $a_0, a_1, a_2, . . . ,$ what is $a_n$? I have a lot of experience going from sequence to generating function, but I am not completely sure how to go the other way. Any guidance would be phenomenal!
2026-04-24 02:11:19.1776996679
If $\sqrt[3]{1 − 2x}$ is the generating function of a sequence $a_0, a_1, a_2, . . . ,$ what is $a_n$?
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You should have $(\sum_{i=0}^\infty a_ix^i)^3=1-2x$. This immediately implies $a_0=1$. Looking at the coefficient on the left of $x$, we get $a_0^2a_1+a_0a_1a_0+a_1a_0^2=3a_1=-2$, so $a_1=-2/3$. For the coefficient of $x^2$ you get $a_0a_0a_2+a_0a_2a_0+a_2a_0a_0+a_0a_1a_1+a_1a_0a_1+a_1a_1a_0=3a_0a_2+3a_1^2a_0=0$ or $3a_2+4/3=0, a_2=-4/9$, and so on. For example the coefficient of $x^3$ is $$0=a_0a_0a_3+a_0a_3a_0+a_0a_0a_3+a_0a_1a_2+a_1a_0a_2+a_2a_0a_1+a_2a_1a_0+ a_0a_2a_1+a_1a_2a_0 +a_1a_1a_1.$$ So $0=3a_3+6*8/27-8/27$, $a_3=-\frac{40}{81}$.