Scans are never constant-folded

Description

import pytensor
import pytensor.tensor as pt

x0 = pt.zeros(())
xs, _ = pytensor.scan(lambda x: x+1, outputs_info=[x0], n_steps=4)
fn = pytensor.function([], xs)
fn.dprint()  # Scan still in the graph

This happens because Alloc never constant-folds if used by a SetSubtensor, as most times we want to write in place (and we can't write in-place of constants). But when the whole chain could ultimately be constant-folded (as here), this is wasteful.

The logic for whether to constant-fold based on the graph or not should be the responsibility of the constant-fold rewrite, not the Op. Right now it's implemented here.

AllocEmpty never constant_folds:

https://github.com/pymc-devs/pytensor/blob/17c675a2d908661f9d1f84ae56f4cb6f5c4fa7c8/pytensor/tensor/basic.py#L4397-L4398

And Alloc has this logic in it:

https://github.com/pymc-devs/pytensor/blob/17c675a2d908661f9d1f84ae56f4cb6f5c4fa7c8/pytensor/tensor/basic.py#L1780-L1818

Start with the Alloc and AllocEmpty implementations in pytensor/tensor/basic.py at the linked sections, then reproduce the provided pytensor.scan example and inspect its fn.dprint() output. Trace the constant-fold rewrite and move the graph-dependent decision out of the Ops; done means the example no longer retains Scan while existing in-place behavior remains correct.