This is not an answer but to long for a comment and possibly useful for numerics.
Lemma: A number $n$ which contains a prime factor $p$ which satisfies $p > \sqrt{n}$ cannot be decomposed into triangular divisors.
Proof: Assume $n=p\cdot q\cdot r$ then $p > \frac{n}{p}=q\cdot r \ge q+r$ if $q,r \ge 2$ so the triangle inequality is not satisfied. The tuple $p, \frac{n}{p}, 1$ doesn't satisfy the triangle inequality either.
I first thought that the converse of this lemma (ie numbers without such prime factors can be decomposed into triangular divisors) holds as well but this is false. It fails the first time for $n=30$.
Edit: Numbers with no prime factor bigger than $\sqrt{n}$ is a sequence in oeis.org. This sequence has density $1-\ln(2)$. As our sequence is a subset of this one, this disproves my intuition of almost all numbers and shows that the sequence has in fact density smaller equal $1-\ln(2)$ (the numerics suggest a density of $0$).
Any transformation is decomposable into translation, deformation (here only isotropic deformation is present, due to equilaterality) and rotation. Rotation is done by three subsequent rotations with respect to 3 different axes, where the transformation order matters. The triangle after transformation is given by formula:
\begin{align}
\left[\begin{array}{c}x_i\\y_i\\z_i\end{array}\right]=\left[\begin{array}{c}X_0\\Y_0\\Z_0\end{array}\right]+
\underbrace{\frac{\text{len}_2}{\text{len}_1}
\left[\begin{array}{ccc}
c_3 & s_3 & 0 \\
-s_3 & c_3 & 0 \\
0 & 0 & 1 \\
\end{array}\right]
\left[\begin{array}{ccc}
c_2 & 0 & s_2 \\
0 & 1 & 0 \\
-s_2 & 0 & c_2 \\
\end{array}\right]
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & c_1 & s_1 \\
0 & -s_1 & c_1 \\
\end{array}\right]}_{\mathbf{A}}
\left[\begin{array}{c}a_i\\b_i\\0\end{array}\right],~~i=1,2,3,
\end{align}
where $\text{len}_2<\text{len}_1,~z_i>0,~c_i=\cos(\theta_i),~s_i=\sin(\theta_i),~i=1,2,3$, $~\theta_1=\theta_x,~\theta_2=\theta_y,~\theta_3=\theta_z$, where matrix $\mathbf{A}$ is known and where the vector $[X_0,~Y_0,~Z_0]^{\text{T}}$ represents translation.
Continuation:
Moreover we have known distances $L_1,L_2,L_3$ so that we want
\begin{align}
&(x_i-a_i)^2+(y_i-b_i)^2+z_i^2=L_i^2,~~i=1,2,3,
\end{align}
and
\begin{align}
&z_i>0,~~i=1,2,3.
\end{align}
Also we have 12 equations
\begin{align}
&\left[\begin{array}{c}x_i\\y_i\\z_i\end{array}\right]=\left[\begin{array}{c}X_0\\Y_0\\Z_0\end{array}\right]+
\mathbf{A}\left[\begin{array}{c}a_i\\b_i\\0\end{array}\right],~~i=1,2,3,~~~~~~~~~~~~~~~(1)\\
&(x_i-a_i)^2+(y_i-b_i)^2+z_i^2=L_i^2,~~i=1,2,3,~~~~~~(2)\\
\end{align}
for 12 unknowns $X_0,Y_0,Z_0,$ $x_i,y_i,z_i,i=1,2,3$. So we can hope, at least in some cases, in a unique solution. If this solution satisfies $z_i>0,~i=1,2,3$ then we are done. The following is however clear:
- There are such input parameters, that problem (1,2) has no solution
- In case $\text{len}_2=\text{len}_1$ there are such input parameters, that problem (1,2) has infinitely many solutions
Proof: 1. for $\mathbf{A}=\mathbf{0},~L_1=L_2=L_3=0$ and different points $P_1,~P_2,~P_3$ the problem definitely has no solution. 2. for $\mathbf{A}=\mathbf{I},~L_1=L_2=L_3=1$ we see that for any $X_0,Y_0,Z_0$ s.t. $X_0^2+Y_0^2+Z_0^2=1$ we get solution, which satifies Eq.(2), ergo we have infinitely many different nontrivial solutions in this case. $\clubsuit$
Remark (uniqueness in case len2<len1): If we insert Eqs.(1) into Eqs.(2) we can solve instead of 12 equations only 3 nonlinear equations:
\begin{align}
&(X_0+(a_{11}-1)a_i+a_{12}b_i)^2+(Y_0+a_{21}a_i+(a_{22}-1)b_i)^2+\\&(Z_0+a_{31}a_i+a_{32}b_i)^2=L_i^2,~~i=1,2,3.~~~~~~(3)\\
\end{align}
This is a problem of intersection of three spheres in $\mathbb{R}^3$, which has in general $0,1,2$ solutions (infinite number of solutions are not possible in case len2<len1). If we can proof, that in case of 2 different solutions one yields negative values of $z_i$ (which is indicated by all numerical solutions, which I have tried), then the proof of uniqueness is made. $\diamondsuit$
Best Answer
$T_n=\dfrac{n(n+1)}{2}$
$\sum_{k=0}^n T_{2k+1}=\sum_{k=0}^n(2k+1)(k+1)$
Expand, and use the fact that $\sum_{k=0}^nk^2=\dfrac{k(k+1)(2k+1)}{6}$ and $\sum_{k=0}^nk=\dfrac{k(k+1)}{2}$