let t(n) denote the number of integer sided triangle with distinct sides chosen from(1,2,3...n) then t(20)-t(19) equals
Go here: The On-Line Encyclopedia of Integer Sequences™ (OEIS™)
Enter sequence number A002623
I believe your question is very similar to this one
http://www.mathhelpforum.com/math-he...es-148807.html
Ignore my initial comments where I misinterpreted what was being asked.
The answer to the OP's question is the number of triangles with integer sides and at least one side 20. The sum of the other two sides can vary from 21 to 40. So we can have 10 + 2* (11 + 12 + ... + 19 ) + 20 = 10 + 2*135 + 20 = 300.
EDIT : Sorry, overlooked the distinct sides part. The number of such isosceles triangles is 9 + 20 = 29. So the answer should be 300 - 29 = 271.
Hmm I think your number is too high because for example wouldn't you get the 19 in parentheses by considering these triangles?
{20,1,38}
{20,2,36}
...
{20,19,19}
and
{20,1,39}
{20,2,37}
...
{20,19,20}
But a side length cannot be greater than 20.
I get 81 as explained in post #25 here, which I probably should have referenced before.