A010815 convolution inverse?
Dean Hickerson
dean at math.ucdavis.edu
Wed Jun 12 08:49:36 CEST 2002
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
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