Can we find lattice polyhedra with faces of area 1,2,3,...? I asked this question two months ago on MSE, where it earned the rare
Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days.
Perhaps justifiably so! Here I repeat it with slight improvements.

Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$,
all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere.
Let $A(P)$ be the area sequence, the sorted list of areas of $P$'s
faces. For example:

 
 
 
 
 
 


Using regular expression notation, this sequence can be written
as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with
$A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e.,
I conjecture there are no golyhedra. Q1. Can anyone prove or disprove
this?
Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more
$a$'s.
For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:

 
 
 


Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$?
The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$,
but end effects destroy the regularity.
The broadest question is: Q3. Which sequences $A(P)$ are achievable?
Can they be characterized? Or at least constrained?

Update (30Apr14). Q1 and Q2 are answered by Adam Goucher's 
brilliant example that achieves
$1^1 2^1 3^1 \cdots 32^1$. In light of this advance,
a more specific version of Q3 may be in order: 
Q3a: Identify some sequence that is not realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron,
described on 
Adam Goucher's blog.
And later a 12-face golyhedron.
 A: By using the Euler characteristic $\chi = L-E+V$ of the graph $\Gamma$ corresponding to the polyhedron (each cube is a vertex and we link adjacent cubes). One can show that 
$$ S = 4N - 2L + 2\chi $$
where $S$ is the total area of the uncovered faces, $N$ the number of cubes and $L$ the number of loops in $\Gamma$. 
This gives a constraint on the possible $A(P)$. For example $A(P)=1^n 2^n 3^n \dots p^n$ can only be constructed if $4\mid np(p+1)$. 
For the interesting case of golyhedra ($n=1$), our constraint reduces to $p \equiv 0,3 \:\mathrm{mod}\: 4$.
$p$-face golyhedra have been constructed by Alexey and Adam for $p=12, 15, 32$. Our constraint suggests that $p=11$ should also be possible. And indeed, we have found an 11-face golyhedron (see image). Because $p=7,8$ are easily ruled out, this has to be the smallest golyhedron.
We are led to conjecture that a $p$-face golyhedron exists if and only if $p \equiv 0,3 \:\mathrm{mod}\: 4$.
A: I found a 32-face example with face areas $\{ 1, 2, \dots, 32 \}$:

It took a reasonable amount of experimentation to stop it from self-intersecting.
