A010815 convolution inverse?

[Christian G.Bower]

Dean Hickerson <dean at math.ucdavis.edu> wrote:
...
> Aren't you confusing "inverse" with "square root"?  A010815 and A000041
*are*
> convolutional inverses, since the product of the corresponding Taylor
series
> is 1.

We should probably avoid the phrase "convolutional inverse" altogether
since there is that potential to interpret "inverse" as reciprocal or as
"undoing the operation."