## Re: [PrimeNumbers] Squares & primes1- At least 2 primes between adjacent squares!!?

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• One of these days I ll learn to ignore these posts, but until then... It s bad enough working out what the mirror sequence is. It s impossible from the way
Message 1 of 5 , Jan 1, 2004
One of these days I'll learn to ignore these posts, but until then...

It's bad enough working out what the mirror sequence is. It's impossible
from the way you've defined it, I had to look at the examples instead.

So you take all the odd numbers between (2n)^2 and (2n+1)^2, of which
there are of course 2n. You then write a sequence of pairs:

{(2n)^2+1, 2n-1},
{(2n)^2+3, 2n-3},
{(2n)^2+5, 2n-5},
...
{(2n)^2+2n-3, 3},
{(2n)^2+2n-1, 1},
{(2n+1)^2-2n, 1},
{(2n+1)^2-2n+2,3},
...
{(2n+1)^2-4, 2n-3}
{(2n+1)^2-2, 2n-1}

You then claim that, of the 2n terms in this sequence, 2n-2 of them
consist of 'actual divisors'? Meaning what? What divides what? It's not
clear.

Andy
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