[seqfan] Re: A214089

[franktaw at netscape.net]

Jonathon,

If p is an odd number, prime or not, p^2 == 1 (mod 8). n# is divisible 
by 2, but not by 4; so (p^2-1) / n# is divisible by 4. [Please note the 
parentheses; what you wrote is equivalent to p^2 - (1 / n#).]

Franklin T. Adams-Watters

-----Original Message-----

I observed that for the first 14 terms in A214089
<https://oeis.org/A214089> , the following holds:

   p^2 - 1 / n# = 4x.

In other words, p^2 - 1 / n# is congruent to 0 MOD 4.

Subsequent to this observation , two new terms were added and the above
holds true for those as well.

Solving for x gives the sequence {1, 1, 1, 1, 19, 17, 1, 2567, 3350,
128928, 3706896, 1290179, 100170428, 39080794, 61998759572, 7833495265}.

Can someone far more familiar with prime numbers explain why this may 
or
may not be true for all a(n)?  I would like to add a comment to the
sequence noting this observation, but I am unsure whether it is in fact
true for all a(n).

  I don't know if this is relevant, but I found a comment, by Robert G.
Wilson, in A118478 <https://oeis.org/A118478> which defines another
sequence whose first seven terms are {1, 1, 1, 1, 19, 17, 1} and also
has 39080794 as its 14th term.

-Jonathan

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