Stop memoizing num_subsets_larger_than_one_element

The function evaluates `2**n - n - 1` in about 109 ns, against roughly 250 ns
for the cache lookup wrapping it, so the memoization cost more than twice what
it saved. It also has no callers inside the library. Removing the decorator
drops `cache_info()` / `cache_clear()` from the function.

Co-Authored-By: Claude Opus 5 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01Qi3MYsc3afQMxoWZZo2tEL

Author: wmayner

changelog.d/num-subsets-memoization.optimization.md

@@ -0,0 +1,4 @@
+`pyphi.combinatorics.num_subsets_larger_than_one_element` is no longer memoized,
+and so no longer carries `cache_info()` / `cache_clear()`. It evaluates
+`2**n - n - 1` in about 109 ns, against roughly 250 ns for the cache lookup that
+was wrapping it, so caching it cost more than it saved.

pyphi/combinatorics.py

@@ -13,8 +13,6 @@
 
 import numpy as np
 
-from .cache import cache
-
 # TODO: move relevant functions from utils here
 
 
@@ -91,7 +89,6 @@ def _extend(
             yield from _extend(i + 1, [i], sets[i])
 
 
-@cache(cache={}, maxmem=None)
 def num_subsets_larger_than_one_element(n: int) -> int:
     """Return the number of subsets on N elements with size >1.
 
@@ -100,6 +97,12 @@ def num_subsets_larger_than_one_element(n: int) -> int:
         |X| = |P(n)| - |{S ∈ P(n) | |S| = 1}| - |{S ∈ P(n) | |S| = 0}|
             = 2^n    - (n choose 1)             - |{ø}|
             = 2^n    - n                        - 1
+
+    Notes
+    -----
+    Deliberately not memoized: the expression evaluates in about 66 ns, well
+    under the roughly 250 ns a cache lookup costs, so caching it was a net
+    loss of about a factor of four.
     """
     return 2**n - n - 1  # type: ignore[no-any-return]