Define the uncentered maximal function $$\widetilde{M}f(n)=\sup_{s,r\in\mathbb{Z}^{+}}\dfrac{1}{s+r+1}\sum_{k=-r}^{k=s}\left|f(n+k)\right|,$$ where $\mathbb{Z}^{+}=\left\{0,1,2,\ldots\right\}$. Define the left and right maximal functions $M_{L}f$ and $M_{R}f$, respectively by $$M_{L}f(n)=\sup_{r\in\mathbb{Z}^{+}}\dfrac{1}{r+1/2}\left\{\dfrac{\left|f(n)\right|}{2}+\sum_{k=-r}^{k=-1}\left|f(n+k)\right|\right\}$$ $$M_{R}f(n)=\sup_{s\in\mathbb{Z}^{+}}\dfrac{1}{s+1/2}\left\{\dfrac{\left|f(n)\right|}{2}+\sum_{k=1}^{k=s}\left|f(n+k)\right|\right\}$$
In the paper "Discrete Tanaka's Theorem", the authors claim that $$\widetilde{M}f(n)=\max\left\{M_{L}f(n),M_{R}f(n)\right\}$$ It is easy to see the direction $\widetilde{M}f(n)\leq\max\left\{M_{L}f(n),M_{R}f(n)\right\}$, but I am at a bit of a loss for the reverse inequality.
Indeed, suppose $f$ is the compactly supported function $$f(n)=\begin{cases} 1 & {n=0} \\ 2 & {n=1} \\ 0 & {\text{otherwise}}\end{cases}$$ Then $M_{L}f(0)=1$ and $M_{R}f(0)=5/3$, but $\widetilde{M}f(0)=3/2$. Am I missing something?
Yes, the word obvious and mistaken claims are often in close proximity of each other. Your counterexample is valid; the stated identity is false.
It holds in the continuous case (and was used by Tanaka), but in the discrete case the central point with its weight $1/2$ throws a monkey wrench into the argument.
By the way, it seems that the conjecture that motivated the paper has been settled in the original, continuous formulation by Ondřej Kurka in On the variation of the Hardy-Littlewood maximal function. Although the sharp constant $C=1$ remains conjectural.