User blog comment:ArtismScrub/Generalization of chained arrow notation/@comment-30118230-20180217152738/@comment-30118230-20180217164218

Not really. If I get your notation correctly,then:

$$a \rightarrow b = a^b$$

$$\#_1 a\rightarrow b \#_2 = \#_1 (\#_1 a-1\rightarrow b \#_2)\rightarrow b-1\#_2$$

$$\#_1 \rightarrow 1 \#_2 = \#_1$$

$$\# a \odot\rightarrow b = \# a \odot a \odot ..... a \odot a$$ with $$b$$ many $$a$$s

$$\# a \odot\rightarrow_{k+1} b = \# a\odot\rightarrow^{b}_k a$$ where $$\odot^1 = \odot \land \odot^{k+1}=\odot\odot^k$$

$$"\rightarrow" := "\rightarrow_1"$$

Where $$\#$$ is either empty or some finite string of integers,seperated by operators and $$\odot$$ is either empty or some finite string of operators.

This definition should cover all valid cases for the notation.