[seqfan] Re: Sequence related to A180919

[Robert Israel]

I'm not "heavy duty" by any means, but if n is even (say n = 2*m),
x^2 = k^2 + 2*m*k + 1 becomes
(k+m+x)(k+m-x) = m^2 - 1
and the number of nonnegative integer solutions is the number of ways to 
write m^2 - 1 as the (unordered) product of two nonnegative integers that 
are either both odd (if m is even) or both even (if m is odd).  For example, with 
n=10, m=5, 24 = 2*12 = 4*6 corresponding to the solutions 
(k=2,x=5) and (k=0,x=1), while with n=16, m=8, 63 = 1*63 = 3*21 = 7*9
corresponding to the solutions (k=24,x=31), (k=4,x=9) and (k=0,x=1).

If n is odd (say n=2*m+1), x^2=k^2+(2*m+1)*k+1 becomes
(2*k+2*m+1-2*x)*(2*k+2*m+1+2*x) = (2*m+3)*(2*m-1)
and solutions correspond to ways to write (2*m+3)*(2*m-1) as the
product of two nonnegative odd integers.  For example, with
n=11, m=5, 117 = 1*117 = 3*39 = 9*13 corresponding to (k=24,x=29),
(k=5,x=9) and (k=0,x=1).

Robert Israel                                israel at math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            Vancouver, BC, Canada

On Tue, 18 Jan 2011, Joerg Arndt wrote:

> (If I understood the message correctly)
> I'd suggest to just create one sequence
> "Number of squares of the form k^2+n*k+1 where k>0."
>
> But first wait for comments from the heavy duty number theorists.
>
> * Bruno Berselli <berselli.bruno at yahoo.it> [Jan 18. 2011 17:10]:
>> [...]
>
>
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