[seqfan] Re: help with sequences of form a(n) = Product_{i=1..n} j^i - k^i

Bob Selcoe rselcoe at entouchonline.net
Sun Mar 6 19:40:08 CET 2016

Hi Jean-Paul,

Thanks - very clear.  It well explains q-Pochhammer symbols and why G.C. 
Greubel's formulas apply. However, it doesn't explain why q=2..12 in the 
xrefs for those entries; for example, in xrefs for A027637, we see Cf. 
A027871 (q=3) and A027872 (q=5), among others.  But in the formulas for 
those entries, a=q = 1/3 and 1/5, respectively.

I think the meaning of q needs to be clarified, or "(q=x)" removed from the 
xrefs in all the entries.  Perhaps instead of q=2,3,4 etc. in the xrefs,  it 
could show a=q = 1/2, 1/3, 1/4 etc.?

re: limit ratios - another way of stating my question is: when a=q>0, does 
(a;q)_{\infty} = 0 for any {a;q}??   I initially thought a,q >= ln(2), but 
now I think there is no solution (unless I'm mistaken, a,q < 1 by 
definition, and (a;q)_{\infty} = 0 iff a,q = 1).


From: "jean-paul allouche" <jean-paul.allouche at imj-prg.fr>
Sent: Sunday, March 06, 2016 3:22 AM
To: <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: help with sequences of form a(n) = Product_{i=1..n} 
j^i - k^i

> Dear Bob
> My two cents: is the following clear enough?
> (a;q)_n = (1-a) (1-aq) (1-aq^2)...(1-aq^{n-1}).
> As usual if n=0, the above product is empty and
> is considered as equal to 1.
> The case where n is infinite, noted (a;q)_{\infty}
> is just the "limit" of (a;q)_n when n goes to infinity.
> best
> jean-paul
> Le 05/03/16 20:24, Bob Selcoe a écrit :
>> q-Pochhammer symbol
> --
> Seqfan Mailing list - http://list.seqfan.eu/

More information about the SeqFan mailing list