Duplicate hunting, pt. 10

[Andrew Plewe]

submissions to allow for list members to take care of as many as possible).
submitted as well; I can post that on the blog but I don't think it'd be of
	-Andrew Plewe-
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> For ease of setup and speed, I've created a blog:
> http://duplicate-sequences.blogspot.com/

Great idea. A wiki would be even better, aren't there some free
providers for that ?

A question close to the topic: what about sequences that just differ by 1,
i.e. a(n)=b(n)+1 for all n ?
I ask it since I just ran across the following (included below), the
comment is easily proved:
A117805(n+1) = (A117805(n)-1) A117805(n)  by definition, and
A005267(n+1) +1 = product(A005267(k), k=1..n ) by definition.
If, by assumption, the latter is equal to A117805(n),
then A117805(n+1) = A005267(n+1) product(A005267(k), k=1..n )
 =product(A005267(k), k=1..n+1 ) = A005267(n+2) +1,
i.e. the assumption is true for all following n.

M.H.

http://www.research.att.com/~njas/sequences/A117805 :
A117805 		Start with 3. Square the previous term and subtract it. 	
	3, 6, 30, 870, 756030, 571580604870, 326704387862983487112030,
106735757048926752040856495274871386126283608870,
11392521832807516835658052968328096177131218666695418950023483907701862019030266123104859068030
(list; graph; listen)
	
	COMMENT
Apparently a(n)=A005267(n+1)+1. - R.J. Mathar
(mathar(AT)strw.leidenuniv.nl), Apr 22 2007