The idea is to use induction on $n$. For $n \in \{1;2\}$ thesis holds. Lemma 1: \[ T_{2n}=\left[{\frac{17}72^{n-1}}\right]-1 \longrightarrow T_{2(n+1)-1}=\left[{\frac{12}72^n}\right] -1 \] proof We obviously know that \[\left[{2\frac572^{n-1}}\right]=\left[{2\frac572^{n-1}}\right] \] \[ \left[{\frac{10}72^{n-1}}\right]=\left[{\frac572^n}\right] \] \[ \left[{2^{n-1}+\frac372^{n-1}}\right]=\left[{\frac572^n}\right] \] Whereas $2^{n-1}$ is integer, we can write it outside the function of integer part without changing something \[ 2^{n-1}+\left[{\frac372^{n-1}}\right]=\left[{\frac572^n}\right] \] \[ 2^{n-1}+\left[{2\cdot2^{n-1}+\frac372^{n-1}}\right]=2^n+\left[{2^n+\frac572^n}\right] \] \[ 2^{n-1}+\left[{\frac{17}72^{n-1}}\right] -1 = \left[{\frac{12}72^n}\right] -1 \] Since we assumed that \[ T_{2n}=\left[{\frac{17}72^{n-1}}\right]-1 \] we have \[ T_{2n} + 2^{n-1} = \left[{\frac{12}72^n}\right] -1\] and by hypothesis we have \[ T_{2(n+1)-1}=T_{2n}+2^{n-1} \] And so \[ T_{2(n+1)-1}=\left[{\frac{12}72^n}\right] -1 \] Q.E.D. Lemma 2 \[ T_{2(n-2)-1}= \left[{\frac{12}72^{n-3}}\right] -1 \longrightarrow T_{2n}= \left[{\frac{17}72^{n-1}}\right] \] proof We obviously know that \[ \left[{4\frac372^{n-3}}\right] = \left[{4\frac372^{n-3}}\right] \] \[ 2^n+\left[{\frac{12}72^{n-3}}\right] = 2^n+\left[{\frac372^{n-1}}\right] \] \[ 2^n+\left[{\frac{12}72^{n-3}}\right] = \left[{2\cdot2^{n-1}+\frac372^{n-1}}\right] \] \[ 2^n+\left[{\frac{12}72^{n-3}}\right]-1 = \left[{\frac{17}72^{n-1}}\right]-1 \] Since we assumed that \[ T_{2(n-2)-1}=\left[{\frac{12}72^{n-3}}\right] -1 \] we have \[ T_{2n-5} +2^n = \left[{\frac{17}72^{n-1}}\right]-1 \] by hypothesis \[ T_{2n}=T_{2n-5}+2^n \] so \[ T_{2n} = \left[{\frac{17}72^{n-1}}\right]-1 \] Q.E.D. Using alternately lemma 1 and lemma 2 we need two consecutive value of $n$ to establish that thesis holds for every following value of $n$, and since for $n=1$ and $n=2$ thesis holds we have finished. P.S. If there is something not very clear please tell me.