[extra__tofu]

   n=1, A=2:
   h    b   #tri  #non-congruent tri
   1    4   b-1   roundup(#tri/2)
   2    2   b-1   roundup(#tri/2)

   .
   .
   .
   
   n=4, A=16:
   h    b      #tri  #non-congruent tri
   1    2A/h   b-1   roundup(#tri/2)
   2    2A/h   b-1   roundup(#tri/2)
   4    ...    ...
   8    ...
   16   ...          ...

summing up #non-congruent tris for each n I see

   n  tri
   1  3
   2  7
   3  15
   4  31

so I think that answer may be 2^(n+1) - 1.

Could be wrong because I forgot what congruent really means.

[tylerhou]

You're undercounting some triangles and missing the acute condition. For n = 4, there is more than one triangle with base 4 and height 4, for example: [(0, 0), (4, 0), (1, 4)]; [(0, 0), (4, 0), (2, 4)]; (and [(0, 0), (4, 0), (3, 4)] but it's congruent to the first so we don't count it). Furthermore, a triangle with base 16 and height 1 might be non-acute.

[dilatedmind]

for the first 8 im getting 1,3,6,14,28,60,120,248

[yantrams]

Close enough but wrong :) I went the same route as you did and realized my mistake after a while. Base and altitude values can be fractional/irrational too.

PS : I bruteforced my way to the answer.