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Using generating functions on the first one gets a bit messy; since it’s first order, you can simply ‘unwind’ the recurrence. In what follows I use the notation $$x^{\underline k}=\underbrace{x(x-1)\ldots(x-k+1)}_{k\text{ factors}}$$ for the falling factorial power. Shifting the index by $1$, we see that the recurrence can be written $i_n=2ni_{n-1}+2$. ...
$$e^{-1}+4e^{-9}+9e^{-25}+16e^{-49}\approx 0.36837308051278053458657911933771842$$ should be good enough. The truncation error is $$\sum_{i=4}^\infty(i+1)^2e^{-(2i+1)^2}=\sum_{i=4}^\infty(i+1)^2e^{-4i^2-4i-1}<\sum_{i=4}^\infty(i+1)^2e^{-64-4i-1}<2\cdot10^{-34}$$ as can be computed analytically. You can obtain this estimate from \sum_{i=n}^\infty ...