I am bearing seeking for a sequential decomposition of the norm in Orlicz space. Let me state what is known in the particular case of Lebesgue space $L^p(\Bbb R^d)$.
Given $u\in L^p(\Bbb R^d)$ let $$n\in \Bbb Z,\qquad D_n= \{x\in \Bbb R^d: 2^n \leq |u(x)|<2^{n+1}\},\quad \text{put}\quad d_n= |D_n|= \int_{D_n}d x $$
Then it is not difficult to show that \begin{align}\tag{1}\label{eq1}\|u\|_{L^p(\Bbb R^d)}= \Big(\int_{\Bbb R^d} |u(x)|^pd x\Big)^{1/p}\sim \Big(\sum_{n\in \Bbb Z} 2^{np}d_n\Big)^{1/p}\end{align}
To be more precise, we have $$\frac12 \|u\|_{L^p(\Bbb R^d)}\leq \Big(\sum_{n\in \Bbb Z} 2^{np}d_n\Big)^{1/p}\leq \|u\|_{L^p(\Bbb R^d)}$$
Question is there a way to get a decomposition similar to \eqref{eq1} for an Orlicz space $L^\phi(\Bbb R^d)$? Where $\phi$ is a sufficiently nice Young function, e.g., $\phi$ is continuous, increasing, convex and in addition the mapping $x\mapsto \frac{\phi(x)}{x}$, $x>0$ is increasing and satisfies
\begin{align*} &\lim_{x\to 0^+}\frac{\phi(x)}{x}= \lim_{x\to \infty}\frac{x}{\phi(x)}= 0. \end{align*}
Recall that: \begin{align*} L^\phi(\Bbb R^d)&= \Big\{u: \Bbb R^d\to \Bbb R\text{ meas.}:~ \int_{\Bbb R^d} \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x<\infty ~~\text{for some $\lambda>0$}\Big\}. \end{align*} The space $L^\phi(\Bbb R^d)$ is equipped with the Luxemburg norm $\|\cdot\|_{L^\phi(\Bbb R^d)}$ by \begin{align} \|u\|_{L^\phi(\Bbb R^d)}=\inf \Big\{ \lambda>0~: \int_{\Bbb R^d} \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x\leq 1\Big\}. \end{align}
Define the weighted Orlicz sequence norm $\lVert\cdot\rVert_{\ell_{\Phi,\mathbf{a}}}$ where $\mathbf{a}=(a_k)_{k\in\mathbb Z}$ is a non-negative bounded sequence, by
$$ \lVert (c_{k})_{k\in\mathbb Z}\rVert_{\ell_{\Phi,\mathbf{a}}} =\inf\left\{\lambda>0, \sum_{k\in\mathbb Z} \Phi\left(\frac{c_k}{\lambda}\right)a_k\leqslant 1\right\}. $$ Using the inequalities $$ 2^n\mathbf{1}_{D_n}\leqslant \lvert u\rvert\mathbf{1}_{D_n}<2^{n+1}\mathbf{1}_{D_n}, $$ one gets that $$ \sum_{n\in\mathbb Z}2^n\mathbf{1}_{D_n}\leqslant \lvert u\rvert \leqslant 2\sum_{n\in\mathbb Z}2^n\mathbf{1}_{D_n}. $$ Moreover, $\left\lVert \sum_{n\in\mathbb Z}2^n\mathbf{1}_{D_n}\right\rVert_{\Phi}=\lVert (2^{k})_{k\in\mathbb Z}\rVert_{\ell_{\Phi,\mathbf{d}}}$ where $\mathbf d=(d_k)_{k\in\mathbb Z}$.