[seqfan] Re: A260709 [was: Re: A081650]

[Chris Thompson]

In the context of David Wilson's question

>> Does A260709(n) = A260709(n - 1) for all composite n > 4?

[n > 8 would be a better guess!]

on Nov 19 2015, Emmanuel Vantieghem wrote:

>Sure it is.  If m can be written as a product of two relatively prime
>numbers u and v, then, a(m-1) is a quadratic residu mod k^2 for k = 1 to
>m-1.  Thus, a(m-1) is quadratic residu mod u^2 and mod v^2, thus a(m-1) is
>quadratic residu mod (u*v)^2 and thus, a(m) = a(m-1).

This is fine.

>                                                     If m is p^s for some
>prime p and s > 2, then a(m-1) is quadratic residu modulo p^2t  for t = 1,
>..., s-1. Thus, a(m-1) is quadratic residu mod every power of  p, thus,
>a(m-1) is quadratic residu mod m^2 and hence, a(m-1) = a(m)

This doesn't work for n=4 [a(4)=52 > a(3)=13] or n=8 [a(8)=1009 > a(7)=436}.
It would be true if a(m-1) was coprime to p [and also 1(mod 8) if p=2]
but although that is true for the rest of the a(p^s-1) values calculated by
Robert Israel (n<=70), it isn't obvious why it should be true in general.

-- 
Chris Thompson
Email: cet1 at cam.ac.uk