While analyzing various rapidly growing sequences in pure lambda calculus due to procrastination, I stumbled upon $(n\omega)$ and $(\omega n)$. I found the behavior of these very similar, yet very different sequences interesting, so maybe someone else will, too.

$n$, $\omega$?!

The $\omega$ combinator is defined by $\lambda(0 0)$. It therefore duplicates every term that’s applied. $(\omega \omega)$ will result in an infinite loop.

For this sequence to work, a number $n$ encoded in lambda calculus uses the Church encoding. Church numerals are defined by an $n$-fold composition of the outermost abstraction: $$\langle 2\rangle\equiv\lambda\lambda(1 (1 0))$$ Similar to $\omega$, these terms will also duplicate their first argument – this time $n$ times.

$(n\omega)$

In this reduction $\omega$ gets cloned $n$ times. After that, $\omega$ duplicates itself repeatedly.

$$\begin{align*} (\langle2\rangle\omega)\equiv&\ (\lambda\lambda(1 (1 0))\ \lambda(0 0))\\ \leadsto&\ \lambda(\lambda(0 0)\ (\lambda(0 0)\ 0))\\ \leadsto&\ \lambda(\lambda(0 0)\ (0 0))\\ \leadsto&\ \lambda((0 0) (0 0)) \end{align*}$$

As $\omega$ duplicates itself $n-1$ times, the growth of $(n\omega)$ is in $\mathcal{O}(2^n)$.

For higher $n$, the reduced term will look like this: $$\lambda\underbrace{(((0 0)^2)^2...)^2}_{n-1\ \mathrm{times}}$$

Fun fact: The exact bit count of its binary lambda calculus encoding is $2^{n+2}$.

$(\omega n)$

This term is just the inverted application of $(n\omega)$. Therefore, $n$ first gets duplicated by $\omega$ and then $n$ clones itself $n$ times:

$$\begin{align*} (\omega\langle2\rangle)\equiv&\ (\lambda(0 0)\ \lambda\lambda(1 (1 0)))\\ \leadsto&\ (\lambda\lambda(1 (1 0))\ \lambda\lambda(1 (1 0)))\\ \leadsto&\ \lambda(\lambda\lambda(1 (1 0))\ (\lambda\lambda(1 (1 0))\ 0))\\ \leadsto&\ \lambda(\lambda\lambda(1 (1 0))\ \lambda(1 (1 0)))\\ \leadsto&\ \lambda\lambda(\lambda(2 (2 0))\ (\lambda(2 (2 0))\ 0))\\ \leadsto&\ \lambda\lambda(\lambda(2 (2 0))\ (1 (1 0)))\\ \leadsto&\ \lambda\lambda(1 (1 (1 (1 0))))\\ \equiv&\ \langle4\rangle \end{align*}$$

Due to its repeated internal application, $\omega n$ turns out to represent $n^n$: $$\lambda\lambda\underbrace{1\circ...\circ1}_{n^n-1\ \mathrm{times}}\circ(10)$$

Fun fact: The exact bit count of its binary lambda calculus encoding is $5\cdot n^n + 6$. The factor $5$ is the result of encoding repeated applications of $1$s.

$(\omega_i n)$ and beyond

Adding more applications to the $\omega$ combinator will result in a higher power tower. Using $(\omega_3 n)$ with $\omega_3=\lambda((0 0) 0)$, for example, will reduce to $n^{n^n}$. $(\omega_3\langle 3\rangle)$ already can’t easily be solved by most traditional reducers.

Detour: After getting caught in the depths of Googology, I now know that these power towers have quite interesting notations. Using Knuth’s up-arrow notation, $3^{3^3}$ can be written as $n\uparrow\uparrow3$. One can escalate this even further: $n\uparrow\uparrow\uparrow3$ would equal $n\uparrow\uparrow (n\uparrow\uparrow(n\uparrow\uparrow n))$. Graham’s number abuses this notation to create an unimaginably huge number. Yet, to come back to lambda calculus, numbers even bigger than Graham’s number can be represented in just 8 bytes of binary encoded lambda calculus! See this Googology entry and this codegolf post.

However, these numbers are still basically zero compared to the growth of busy beavers. One of my next posts will discuss some interesting aspects of the busy beaver of lambda calculus, $\mathrm{BB}_\lambda$.

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Methods used in this article include bruijn’s :time and :length commands.