Question about A006906

[David Wilson]

No, I was just wrong. Max Alexeyev was right.

Partitions of 4 are {(4),(3,1),(2,2),(2,1,1),(1,1,1,1)} whose products are 
{4,3,4,2,1} whose sum is 14
Partitions of 3 are 3 => {(3),(2,1),(1,1,1)} whose products are {3,2,1} 
whose sum is 6
Partitions of 2 are {(2),(1,1)} whose products are {2,1} whose sum is 3
Partitions of 1 are {(1)} whose products are {1} whose sum is 1
Partitions of 0 are {()} whose products are {1} whose sum is 1

I guess I forgot about the empty partition. So a(0) is in fact correct.

NJAS: Instead of changing the name, you should keep the old name, which is 
correct, and maybe add a comment for people like me:

%N A006906 a(n) = sum of products of terms in all partitions of n.
%C A006906 0 has one partition (the empty partition) whose product is 1 (the 
empty product) giving A006906(0) = 1.


----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: "Sequence Fans" <seqfan at ext.jussieu.fr>; "David Wilson" 
<davidwwilson at comcast.net>
Sent: Saturday, May 19, 2007 7:42 AM
Subject: Re: Question about A006906


> David is right, strictly speaking, since
> there are no terms.  But the leading term
> seems to want to be 1. So I am
> changing the definition to:
> %N A006906 a(0) = 1; for n >=1, a(n) = sum of products of terms in all 
> partitions of n.
> Neil
>
>> Shouldn't A006906(0) = 3 D 0 ?
>
>
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