John Tromp defines a version of the lambda calculus that is encoded in binary: https://tromp.github.io/cl/cl.html
a) Does there exist a concatenative version of this language (or its combinatory equivalent)?
b) Does there exist a concatenative version restricted to the primitive recursive functions (e.g. as per Combinators for the Primitive Recursive Functions)?
Yes, there exists a version of BLC working similarly to Zot.
Consider the zot implementation at [1], which mimics the original definition at [2]. It defines 3 lambda expressions:
zot = \c.c I
0 = \c.c(\f.f S K)
1 = \c\L.L(\l\R.R(\r.c(l r)))
such that applying zot to a sequence of 0s and 1s produces any desired lambda term. For instance, I = \x.x = zot 1 0 0.
But while it's true that the concatenation of 2 bit sequences A and B is another bit sequence C, the resulting program has little to do with the program for B, since B no longer follows the initial zot term.
The lambda terms for 0 and 1 in zot are very much unlike plain bits.
Is there any way we can use the plain bits 0 = \x\y.x and 1 = \x\y.y instead? It turns out that we can! At the cost of moving all complexity into the initial term. This is what I did in the implementation at [3]. The 195 bit term blc serves the same role as zot above. But by applying it to the sequence of (actual plain) bits that are the BLC code of a lambda term T, we obtain that very term T. For instance, I = \x.x = blc 0 0 1 0
blc is in fact a binary tree containing all closed lambda terms at its leaves, and applying 0/1 is taking the left/right subtree.
It's also possible to reduce the 195 bits to a much smaller 100 bits by using the single point combinator basis A = \x\y\z.x z(y(_.z)) [4].
[1] https://github.com/tromp/AIT/blob/master/ait/zot.lam
As per @cody's comment above, if we require only that "the concatenation of any two programs is a valid program" then I believe that Iota, Jot and Zot (semarch.linguistics.fas.nyu.edu/barker/Iota/zot.html) are examples of a).
However, IMO the more interesting question part of the question is b)...
