[seqfan] Re: The inverse Euler transformation of A035938

franktaw at netscape.net franktaw at netscape.net
Sun Jan 22 22:13:33 CET 2012

Yes, that is correct. That inverse Euler transform follows immediately 
from the first sentence of the definition.

That that and the second sentence of the definition are equivalent is 
what the Andrews reference is about.

Franklin T. Adams-Watters

-----Original Message-----
From: William Keith <william.keith at gmail.com>

On Sun, Jan 22, 2012 at 5:30 PM, N. J. A. Sloane 
<njas at research.att.com>wrote:

> Peter, That's a very nice discovery. Please make sure it gets added
> (with or without a proof) to that sequence.
> It reminds me a bit of the analysis of restricted
> partitions in Comtet's Advanced Combinatorics,
> Section 2.6, pages 108-115. Perhaps that will suggest a proof.
>  Best regards
>                         Neil

What does EULERi do?  If, as I guess, it gives the exponents b_i such 
the generating function of the data sequence is \prod 1/(1-q^k)^{b_i}, 
the 1-0 sequence given is recording Gordon's Theorem itself: entries 7k,
7k+2, and 7k-2 are 0, and others are 1, giving the generating function 
partitions into parts not 7k, 7k+2, or 7k-2.

William Keith


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