<s>[INST] Triangular numbers is relating to Pythagoras triples Let 
$$T_n:=1,3,6,10,15,...$$
are triangular numbers, $T_n={n(n+1)\over 2}.$
Pythagoras triples $(5,12,13),(7,24,25),(9,40,41),(11,60,61)$ and so on,...
Observe that $S_n=T_n+2T_{n+1}+T_{n+2}$ for $n\ge1$
$S_1=13$
$S_2=25$
$S_3=41$ and so on, ...
Can anyone explain how does this formula $S_n$ in term of triangular is related to Pythagoras triple numbers?
 [/INST] Pythagorean triples can be gnerated by $(m^2-n^2, 2nm , n^2+m^2)$ and if $m=n+1$ you will obtain your sequence.
Note that your sequence $S_n$ are the sum of two consecutive squares. $S_=n^2+(n+1)^2$. Also 
\begin{eqnarray*}
S_n&=& \frac{n(n+1)}{2}+ 2\frac{(n+1)(n+2)}{2}+\frac{(n+2)(n+3)}{2}= 2n^2+6n+5 \\&=&(n+1)^2+(n+2)^2
\end{eqnarray*}
again the sum of two consecutive squares (offset by $1$).
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