I also want to know the anwer

Can there be two longest sides? As in a triangle with lengths 1, 2n, 2n? Also what counts as a different triangle. Does a 3, 4, 5 triangle have two different orientations, or if we can reflect one triangle onto another then they are the same?

Start with 1, 2n, 2n. From there you can keep on increasing the first side to 2n, so (1,2n,2n),(2,2n,2n),...(2n,2n,2n). For a total of 2n values Then go to 2, 2n-1, 2n. Keep on increasinng the first side to 2n-1 for a total of 2n-2 values. If we keep repeating this process we get the total sum of different triangles as 2+4+6+...2n, which equals n(n+1)

I guess natmath is right. And, The first one is not very difficult but I don't know the second one.

I find out that $b_{n}=2^{n}n$

Alpha314159 wrote: I find out that $b_{n}=2^{n}n$ Nice, how did you go about finding that out?

Alpha314159 wrote: I find out that $b_{n}=2^{n}n$ Assuming this is true, we have $2^nn\leq2019n(n+1)$ $2^n\leq2019(n+1)$ Try n=15 2^15=32768 2019*16=32304 So the condition is satisfied for 0<n<15, which is 14 values

OH, the answer was wrong, but the idea is right, you can solve $b_{n}$ from the condition, the result is $b_{n}=C_{n}^{1}a_{1}+C_{n}^{2}a_{2}+...++C_{n}^{n}a_{n}$, then , you can calculate the right answer

Alpha314159 wrote: OH, the answer was wrong, but the idea is right, you can solve $b_{n}$ from the condition, the result is $b_{n}=C_{n}^{1}a_{1}+C_{n}^{2}a_{2}+...++C_{n}^{n}a_{n}$, then , you can calculate the right answer How did you get that equation?

first, you guess. when ,n=3, you can brutally solve the equation system \[\left\{ {\begin{array}{*{20}{c}} {C_1^1{b_1} = {a_1}} \\ { - C_2^1{b_1} + C_2^2{b_2} = {a_{\rm{2}}}} \\ {C_3^1{b_1} - C_3^2{b_2} + C_3^3{b_3} = {a_3}} \\ \end{array}} \right.\]and the result is \[\left\{ {\begin{array}{*{20}{c}} {{b_1} = {a_1}} \\ {{b_2} = 2{a_{\rm{2}}} + {a_2}} \\ {{b_3} = 3{a_1} + 3{a_2} + {a_3}} \\ \end{array}} \right.\]secondly, you can prove the general result $b_{n}=C_{n}^{1}a_{1}+C_{n}^{2}a_{2}+...++C_{n}^{n}a_{n}$ by induction or solve $b_{n}$ by the result

I think $b_n=2^{n-2}n(n+3).$