# For which n does the nth “hyperoperation number” n[n]n begin with n in base 2?

If $[n]$ denotes the $n$th binary hyperoperation in the sequence $(+,\times,\uparrow,\uparrow\uparrow,...)$, then the following equality is readily verified for $n=1,2,3,4:$ $$n\,[n]\,n\ =\ (n_2...)_2 \tag{E1}$$ where $(n_2...)_2$ denotes a base-$2$ numeral that begins with the base-$2$ numeral for $n$; thus:

\begin{align} 1\,+\,1&\ =\ (\underline{1}0)_2\\ 2\,\times\,2&\ =\ (\underline{10}0)_2\\ 3\,\uparrow\,3&\ =\ (\underline{11}011)_2\\ 4\,\uparrow\uparrow\,4&\ =\ (\underline{100}...)_2\\ \end{align}

where the RHS of the equation for $n=4$ is a $1$ followed by $2\times 4^{4^4}$ $0$s, because $4\uparrow\uparrow 4=4^{4^{4^4}}=(2^2)^{4^{4^4}}=2^{2\times{4^{4^4}}}$.

Now, except for $n=3$, the above cases $n=1,2,3,4=2^0,2^1,3,2^2$ are all powers of $2$, and it's straightforward to show that E1 holds "trivially" for any $n=2^a$ with $a\ge 1$, because of the following general property of hyperoperators (an easy consequence of their recursive definition):

For any positive integers $x,y,i,j$ with $j\ge i$, there is an integer $c\ge y$ such that $x[j]y = x[i]c.$

In particular, for any $a\ge 1$, there is a $c\ge 2^a$ such that

$$(2^a)[2^a](2^a)=(2^a)[3]c=(2^a)^c=2^{a c}$$

whose base-$2$ numeral is just $1$ followed by $a c$ $0$s. Thus, combining this with the earlier result for $n=2^0$, we have that for any integer $a\ge 0$,

$$(2^a)[2^a](2^a)\ =\ (\underline{10^a}...)_2$$

Now it seems a mere coincidence that E1 holds for $n=3$, and I have no idea how to determine whether it holds for any greater $n$ except for powers of $2$.

NB: Some authors use a different indexing of these same hyperoperations, which we can write as $A_n=[n+1], n=0,1,2,3...$; i.e., such that $(A_0,A_1,A_2,...)=(+,\times,\uparrow,...)$. So it's natural to ask for what $n$ do we now have $$n\,A_n\,n\ =\ (n_2...)_2 \tag{E2}$$

Somewhat surprisingly, in view of the case now for $n=3$, it turns out that this equation holds for $n=0,1,2,3,4$:

\begin{align} 0\,+\,0&\ =\ (\underline{0})_2\\ 1\,\times\,1&\ =\ (\underline{1})_2\\ 2\,\uparrow\,2&\ =\ (\underline{10}0)_2\\ 3\,\uparrow\uparrow\,3&\ =\ (\underline{11}01110111101111001000001110111111110111011)_2\\ 4\,\uparrow\uparrow\uparrow\,4&\ =\ (\underline{100}...)_2\\ \end{align}

For the same reasons as in the earlier equation, this one too must hold when $n=2^a$ for any integer $a\ge 0$, but it seems a slightly more remarkable coincidence that this one, too, holds for $n=3$.

Question: Can it be determined whether E1 or E2 holds for some $n>3$, not a power of $2$?