A010815 convolution inverse?

[Dean Hickerson]

noe at sspectra.com asked:

> A comment says that A010815 is the "convolution inverse of A000041, the
> partition numbers".  What does "convolution inverse" mean?  It does not
> mean the A010815 convolved with itself yields A000041.

Neil replied:

> Normally the conv. inv. of a sequence B is a sequence A such that
> A convolved with itself gives B
...
> In the case of A010815 this was an error, and the correct asserion is:
...
> %C A010815 When convolved with the partion numbers A000041 gives 1, 0, 0, 0, 0, ...

Aren't you confusing "inverse" with "square root"?  A010815 and A000041 *are*
convolutional inverses, since the product of the corresponding Taylor series
is 1.

If A convolved with itself is B, then A is a convolutional square root of B,
not an inverse.

Dean Hickerson
dean at math.ucdavis.edu