Let $T_n$ denote the $n$'th Tribonacci number. They are defined by the difference equation $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ where $T_0=T_1=0$ and $T_2=1$.
Suppose we have two subsequences of the Tribonacci sequence
$$T_{i_0},T_{i_1},\ldots , T_{i_r}$$ $$T_{j_0},T_{j_1},\ldots , T_{j_s}$$
such that $$T_{i_0}+T_{i_1}+\cdots + T_{i_r} = T_{j_0}+T_{j_1}+\cdots + T_{j_s}.$$
My question is, is this equality invariant under shift? In other words, does the following hold $$T_{i_0+1}+T_{i_1+1}+\cdots + T_{i_r+1} = T_{j_0+1}+T_{j_1+1}+\cdots + T_{j_s+1}?$$
I don't know how to prove this, but it looks like it should be easy to prove. If you expand the '$T$'s' by subbing in the formula for $T_n$ you don't get very far. There is probably some trick here that I am not seeing. Any hints would be appreciated!
$T_2=1$, $T_3=1$, $T_4=2$, $T_5=4$, $T_6=7$, $T_7=13$.
This conjecture fails at the first hurdle: \begin{eqnarray*} T_2+T_4+T_5=T_6, \end{eqnarray*} but \begin{eqnarray*} T_3+T_5+T_6 \color{red}{\neq} T_7. \end{eqnarray*}