The boundedness of norms of iterates of a linear operator Let $T$ be a linear bounded operator on a Banach space $X$.
Suppose that $(a_n)$ and $(b_n)$ are two increasing sequences of positive integers such that $b_n > a_n$ for any $n$ and such that $\sup_{n}(b_n - a_n)<\infty$.
Furthermore, assume that $\sup_{n}\|T^{b_n}\|<\infty$.
Does it follow necessarily that $\sup_{n}\|T^{a_n}\|<\infty$?

Note: the assumption clearly implies that the spectral radius of $T$, i.e., $\rho(T) = \lim \|T^n\|^{1/n}$, is at most $1$. However, an operator with spectral radius $1$ can   have unbounded norms of iterates, for example
$T = \left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right) $ on $\mathbb{R}^2$.
 A: No, you may construct a counter-example. But you need to fine-tune the growth.
Denote $N_d$ the standard nil-potent matrix in ${\Bbb R}^{d}$, so e.g.:
$N_3=\left(\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}\right)$. 
Let $\rho_k = 2^{2^k}$ which verifies $\rho_{k+1}=\rho_k^2$ and set $d_k=\rho_k+1$. Also let $\alpha_k = \frac{1}{\rho_k} \log k$.
Our Banach space is the direct sum:
 $$ X = \oplus_{k\geq 1} X_k , \;\; X_k= \ell^{\infty}(\{1,..,d_k\}), $$
equipped with the sup-norm. We consider the diagonal action $A$ of 
$$ A_k = e^{\alpha_k} N_{d_k} : X_k \rightarrow X_k$$
The norm of each $A_k^n$ is either $e^{n\alpha_k}$ if $n\leq \rho_k$ and zero otherwise. As $\alpha_k$ is decreasing it follows that 
 $$ \|A^n\| = \max \{e^{n \alpha_k} :  \rho_k \geq n\}$$
i.e. evaluated for the smallest $k$ for which $\rho_k\geq n$. In particular,
  $$ A^{\rho_k} = e^{\rho_k \alpha_k} = e^{\log k} = k \rightarrow +\infty$$ as $k\rightarrow +\infty$ whereas:
 $$ A^{\rho_k + 1} = e^{(\rho_k+1) \alpha_{k+1}} = \exp(\frac{\rho_k+1}{\rho_k^2} \log k ) \rightarrow 1.$$
In terms of your notation $a_n=\rho_n$, $b_n=\rho_n+1$ and $b_n-a_n=1$ for all $n$.
