[seqfan] Re: Help needed

[Douglas McNeil]

The terms are nearly squares, so (4*n+2)**2 gives values that are almost right:
       5 ,    6,     7,      8,       9,      10,    11
   484,  676,  900, 1156, 1444, 1764, 2116

the difference vector is

   1,  9,  1,  1,  1,  1, 25

Both values which aren't unity have n two more than a square, so maybe
something like

x = (4*n+2)**2 - {1 if (n-2) is not a square,
                          (2*sqrt(n-2)-1)**2 if (n-2) is square}

Or if that's too mundane, the difference vector is a subsequence of
gcd(n, 2^n + 1), so how about

(4*n+2)**2 - gcd(39+n, 2^(39+n)+1)?

<0.94 wink>


Doug

-- 
Department of Earth Sciences
University of Hong Kong