clearvars 

  list=[];

  tot=sym(1/16);

  for n=1:2000

    m=  sym(2/(1/16-3/n));

    if m>0 && m==floor(m) && m<Inf

      disp([m n])

      list(end+1)=m;

      list(end+1)=n;

    end

  end  

  length(list)

  length(unique(list))

>> meansOfPairs

[1568, 49]

[800, 50]

[544, 51]

[416, 52]

[288, 54]

[224, 56]

[160, 60]

[128, 64]

[96, 72]

[80, 80]

[64, 96]

[56, 112]

[48, 144]

[44, 176]

[40, 240]

[38, 304]

[36, 432]

[35, 560]

[34, 816]

[33, 1584]

ans =

    40

ans =

    36

The pairs appear above. The two answers at the bottom are, first, the number of numbers appearing, that is, twice the number of pairs; and, second, the number of distinct numbers. 

 

56, 64, 80, 96 are duplicated, so while 40 numbers appear recorded in the list, there are only 36 unique numbers.

>> mean(list)

ans =

                     244.6

>> mean(unique(list))

ans =

          263.555555555556

                    

so if the duplicated values are counted the two times each appears, the mean is 244.6; if they are counted only once, the mean is 263 + 5/9.