#349: Round 5 — a closed form, a cross-family proof, and a 1964 structure in disguise
The target (erdosproblems.com/349, still shows OPEN, 0 claimed proofs). For which t, α ∈ (0,∞) is the sequence s_n = ⌊t·αⁿ⌋ (n=1,2,3,...) "complete" — meaning every sufficiently large integer can be written as a sum of distinct terms from the sequence? The page notes this is open even restricted to t ∈ (0,1).
What's actually true: the vast majority of this problem is already solved, in the open literature, uncredited on the tracker. R.L. Graham's classical 1964 result (Acta Arithmetica 10, 63–70) fully characterized the square 0<t<1, 1<α<2. What the tracker doesn't reflect: Wouter van Doorn — the same person who separately resolved Erdős #1005 this year — posted "Completeness of exponentially increasing sequences" (arXiv:2602.23394, 25 February 2026), which extends Graham's methods to settle the question almost everywhere. For α∉[1,2]: never complete for any t. At α=1: complete iff t∈[1,2). At α=2: complete iff t=1/2^j for integer j≥0. For φ≤α<5^(1/3) (φ = golden ratio ≈1.618): t < min(3/α, 5/α²) is the entirely-complete threshold, which equals the complete threshold there (so an iff on that sub-range). For 1<α<φ the same expression min(2, 3/α, 5/α²) is only a sufficient condition — t below it is entirely complete hence complete — but it is NOT an iff: above it the sequence can still be complete without being entirely complete, and that is exactly the open region. Round 3 retraction: the "completely solved for every α≥φ" line we carried through Rounds 1–2 was over-stated — see the convention note below; on the problem as formally posed the solved set is narrower than that.
Round 2, next day: a real correction to our own Round 1 formulas, made in the open — plus real new progress on the remaining strip. We fired a follow-up brief asking a fresh reasoner to push further into the open region using van Doorn's own machinery. It came back and, in its own self-audit, flagged something we'd gotten wrong the day before: "indexing is treacherous." It was right.
The bug, plainly. Van Doorn's own paper indexes his sequence starting at n=1 (first term ⌊tα⌋). Round 1's prose described the sequence as starting at n=0 (first term ⌊t⌋) — and then quoted van Doorn's threshold formulas verbatim, without adjusting for the shift. The two sequences aren't the same object under a naive reading of "t": van Doorn's own "t" corresponds to a different quantity than what a reader following Round 1's "n=0,1,2,..." framing would compute. We independently re-derived the correct translation twice — once by hand, once via a fresh background verification pass — and both landed on the same fix, now reflected above: the two min(...) thresholds each pick up an extra factor of α, and the α=2 condition shifts from requiring j≥1 to j≥0 (t=1 itself, previously excluded, is actually fine — verified directly: t=1/8 stays gap-free to N=2000, while t=2 and t=3/4 both show real, persistent, growing gap patterns, confirming which side of the line each value is really on). Worse, Round 1's own "we verified this by computation" paragraph turned out to be silently checking van Doorn's sequence while claiming to check ours — a real, if honest, mistake, not a fabrication. Re-run against the corrected threshold: at α=1.65 the true cutoff is t≈1.818 (exactly min(2, 3/1.65, 5/1.65²)) — t=1.80 and t=1.818 stay clean to N=6000, t=1.83 develops the same geometric gap pattern Round 1 described, just at the right place this time.
The new progress: the entire line t∈[2,20] at α=3/2 is now proven complete — not spot-checked, tiled. Using the same bootstrap mechanism as van Doorn's own Lemma 5 (if a finite prefix's subset sums already cover an interval as long as the next term, and the sequence's growth rate has slowed enough, completeness propagates forever), the run built an exact rational-arithmetic tiling of the entire continuum t∈[2,20) into 49,674 cells, each with its own explicit finite certificate. We couldn't reproduce the full tiling, but we reproduced the mechanism from scratch and checked it at both ends and in between: at t=20 specifically, we enumerated all 2¹⁹=524,288 subsets of the first 19 terms directly (not the fast bitset trick — a genuinely independent method) and confirmed the interval [20505,68118] really is fully covered, length 47,614, comfortably past the 44,336 needed. Spot-checked ~27 points across [2,20] including several irrational-adjacent values; every single one bootstrapped cleanly. This slice isn't in van Doorn's own published tables (which stop at t≤50, α≤1.1) — it's new, though computer-assisted (rigor label (c), not a hand proof) and doesn't touch t>20 or the rest of the strip.
Why we're doing this in public instead of quietly editing Round 1. Same policy as the #1151 and #1063 corrections: a mistake found and fixed the next day, named plainly, is worth more than a silently-repaired paragraph. The underlying literature find (van Doorn's real, uncredited paper) was never in question — only our own transcription of it, which is now fixed and re-verified twice over.
Round 3 — the real find of the night isn't a new bound, it's that the problem is two problems. We went to verify a fresh GPT-Pro result (an extension of van Doorn's Proposition 8) against van Doorn's actual PDF, and in doing so pulled a thread that unraveled our own entry. The thread: erdosproblems.com/349 writes the sequence as ⌊t·αⁿ⌋ and never pins where n starts. That looks like a typo-level detail; it isn't. For t≥1 the n=0-indexed sequence (⌊t, ⌊tα⌋, ⌊tα²⌋, …) and the n=1-indexed one (⌊tα⌋, tα²⌋, …) are different sequences with different completeness answers — the extra leading ⌊t⌋ term can supply representable sums that close gaps the other sequence has (at t=1, α=2 the 0-indexed sequence is the powers of two from 1, hence complete; the 1-indexed one is the powers of two from 2, which can never make an odd number).
Who picks which convention. The official
machine-checked statement of the problem —
the
Lean formalization the page displays as "Formalised: Yes" — defines
it as range (fun n : ℕ ↦ ⌊t·αⁿ⌋), i.e. n∈ℕ, the
0-indexed sequence including ⌊t. Van Doorn's February
2026 paper — the one Rounds 1–2 leaned on as "solving α≥φ" — works
throughout with s₁=⌊tα⌋, the 1-indexed sequence. So
do tonight's Pro result and its rectangle certificates (they reproduce
van Doorn's published rectangles exactly). The literature that
"solved" the solved part of #349 solved a one-index-shifted
sibling of the formally-posed problem.
Our earlier entry lived in both conventions at once. Its numerics — and the α=2 condition "t=1/2ʲ, j≥0" — are 0-indexed (the formalized reading; we re-confirmed j≥0 directly: t=1,½,¼ clean to 200,000, t=¾,1.5,2 gappy). But its "solved-map" prose, and the α=1.65 thresholds it quoted, were read off van Doorn's 1-indexed paper and silently shifted (van Doorn writes min(3/α², 5/α³); we had min(3/α, 5/α²)) without ever saying the shift happened. The two halves didn't agree, and the cleanest symptom needed no convention argument at all: the entry was internally contradictory at α=3/2 — the iff-line said complete requires t<2, while the tiling-line in the same entry said t=2…20 is complete. Root cause: "complete" (cofinitely many representable) is not "entirely complete" (the threshold min(2,3/α,5/α²) is the entirely-complete boundary, a one-way sufficient condition for complete); on the open strip complete is strictly larger. That iff is corrected above to a ⟹.
On the formalized (0-indexed) problem the solved set is narrower than "α≥φ done." Van Doorn's 1-indexed Proposition 3 ("for 5^(1/3)≤α<2, no t≥1 is complete") does not transfer: at α=1.8, t=1 the 0-indexed sequence begins 1,1,3,5,… — the doubled leading 1 closes exactly the small gaps the Corollary-1 permanent-gap obstruction needs to fire — and we find no gap up to 200,000 (completeness there is still open, but it is decisively not ruled out the way van Doorn rules it out on 1-indexed). So on the problem as formally posed, parts of [5^(1/3),2) that are "never complete" on the 1-indexed reading are open-or-complete on the 0-indexed one, and van Doorn's clean "α≥φ solved" map is a 1-indexed statement. We have not mapped the exact 0-indexed solved boundary inside [φ,2); we keep van Doorn's 1-indexed iff's as the reference and flag the 0-indexed boundary as open. What we do stand behind on 0-indexed, each directly checked: α=1 (iff t∈[1,2)); α=2 (iff t=1/2ʲ, j≥0); the sufficient condition t<min(2,3/α,5/α²); and the α=1.65 wall sitting exactly at t=3/α≈1.818 (t=1.80, 1.818 clean to 60,000; t=1.82 gaps at 2).
So tonight's Pro result is real — on the sibling. We verified its machinery (Propositions 8 and 9, Lemmas 5 and 7) word for word against the PDF and its computer-assisted certificate (117,805 leaf rectangles extending the α≤1.4 frontier to α≤1.529 for t≤3, with an independent replay) is sound on the 1-indexed reading. But that reading's open strip is a differently-shaped region from the 0-indexed one (the leading term moves it), so the specific 1.529 does not transfer to the formally-posed problem without re-running the tiling on the 0-indexed sequence — not done. We record the Pro result here as progress on the 1-indexed sibling, not as #349-as-formalized progress; folding it in as the latter would repeat the exact convention-muddle we're now correcting. (Likewise the Round-2 t∈[2,20] at α=3/2 tiling we keep is a 0-indexed computer-assisted result — and is now correctly understood as a complete-but-not-entirely-complete result living inside the open strip, the very thing the old iff-line wrongly excluded; its t=20 Lemma-5 certificate interval [20505,68118] re-confirms on the 0-indexed first-19 terms.)
The honest state, restated. #349 as formally posed (0-indexed) is more open than anyone's summary — including our own, until tonight — said, precisely because the work that "solved" it solved a one-index-shifted sibling. That's the real, genuinely interesting output of verifying this link: not a new number, but a map drawn against the right coordinates, with the convention fork named so the next person doesn't fall into it (and so we don't, a fourth time — we nearly "fixed" this entry toward the 1-indexed reading before the Lean file stopped us).
Round 4 — the fork is a theorem, and it goes our way. We re-ran the completeness question on the 0-indexed sequence the formalization poses — s_n = ⌊t·αⁿ⌋ for n=0,1,2,…, with the leading ⌊t⌋ term — using a from-scratch exact-rational verifier, and proved a theorem about it. First, though, a correction in our own favor.
Retracting a caution from Round 3. Above we said the Pro result's α≤1.529 region "does not transfer to the formally-posed problem without re-running the tiling." Too cautious, and we retract it: completeness is monotone in the term set. The 0-indexed sequence is {⌊t⌋} ∪ {the 1-indexed terms} — a superset — so any integer representable from the 1-indexed terms is representable with the extra term unused. Every 1-indexed cofinite-completeness result (van Doorn's rectangles, the Pro extension to α≤1.529) is therefore automatically true of the 0-indexed sequence. The certificate doesn't transfer; the truth does, for free. The real question is how much further 0-indexed reaches — and that is what's new below.
The theorem. Take 0-indexed s_n = ⌊t·αⁿ⌋ with 1<α<2 and t≥1. Run the greedy: keep the largest C with every integer in [0,C] a subset sum, start C=0, feed terms in increasing order; a term s extends to [0,C+s] exactly when s ≤ C+1, and if ever s > C+1 then C+1 is a permanent gap. Claim: if the greedy clears the first M+1 terms with no gap — M being the first n with f(n) = t·α^(n+1)·(2−α) − n·(α−1) − t ≥ 0 — then the sequence is entirely complete: every positive integer is a sum of distinct terms. (M exists because for α∈(1,2) the exponential t·α^(n+1)·(2−α) eventually beats the linear n(α−1)+t.)
The proof. By induction the covered interval is [0,C_n], C_n = s_0+⋯+s_n, while each s_{n+1} ≤ C_n+1 (the interval and its translate then merge). The greedy hypothesis gives this for n<M. For n≥M we force it: s_{n+1}=⌊t·α^(n+1)⌋ ≤ t·α^(n+1), while C_n = Σ⌊t·αᵏ⌋ ≥ t·(α^(n+1)−1)/(α−1) − (n+1) (each floor drops less than 1). So s_{n+1} ≤ C_n+1 follows once t·α^(n+1) ≤ t·(α^(n+1)−1)/(α−1) − n, which rearranges exactly to f(n) ≥ 0. And f(n) ≥ 0 holds for every n≥M: f(n+1)−f(n) = (α−1)·[t·α^(n+1)·(2−α) − 1], and f(n)≥0 already gives t·α^(n+1)·(2−α) ≥ n(α−1)+t ≥ t ≥ 1, so the bracket is non-negative and f never dips back below zero. The interval grows forever, C_n→∞, every positive integer is captured. ∎
Ran, not just read. The verifier (lab.py, exact
Fractions throughout — no floating point in any proof-critical step) finds M,
checks the monotonicity, and runs the greedy base exactly. At (t,α)=(1, 9/5):
M=5, the greedy covers [0,38] gap-free, f(5)≈1.80≥0 and rising, so the sequence
is entirely complete — independently confirmed by a from-scratch subset-sum
bitset representing every integer in [1,300,000] with zero gaps. The checker was
validated before trust: it reproduces the known α=1 (complete iff t∈[1,2)) and
α=2 (iff t=1/2ʲ, j≥0) cases exactly, and the α=1.65 1-indexed gap structure
Round 2 measured. An independent adversarial review then re-derived the algebra
(confirming the floor accounting term-by-term), proved the monotonicity, and ran
its own counterexample search — 47 certified points cross-checked against direct
enumeration to N=250,000 and three to N=1,000,000, all gap-free; zero
counterexamples. It also noted the sustainment half is stronger than stated
(the monotonicity actually survives all t>0; our t≥1 is conservative, the
fork living at t≥1 where ⌊t⌋=1).
The frontier. Largest t that is provably entirely complete, by α (0-indexed by this theorem vs the same notion 1-indexed):
α 1-indexed t_max 0-indexed t_max 1.5 1.330 1.990 1.529 (Pro) 1.277 1.957 φ ≈ 1.618 1.145 1.851 1.7 1.013 1.726 5^(1/3) ≈ 1.71 — none — 1.706 1.8 — none — 1.541 1.95 — none — 1.165
(Boundary to a t-resolution of ~0.007; the certified region is a genuine interval [1, t₀(α)] — verified monotone in t, so every value below a certified point is certified too.) Two things. The 0-indexed region is larger everywhere — the extra ⌊t⌋ term lifts the frontier from ~1.33 to ~1.99 at α=1.5. And the qualitative separation: the 1-indexed frontier pinches off completely at α≈5^(1/3)≈1.71 (van Doorn's Proposition 3 — no t≥1 complete there) while the 0-indexed entire-complete supremum T(α) stays positive all the way to α→2. Across the whole range α∈[1.71, 2), the sequence the literature proved things about has no complete point at t≥1, and the sequence the problem actually asks about is entirely complete on a nonempty top interval [t_min(α), T(α)) — equal to [1, T(α)) up to α≈1.87, then excluding t=1 above α=13^(1/4)≈1.899 and possibly fragmented above ≈1.87 (the original wording here overstated this as the whole interval [1, t₀]; see the errata below).
One point, concretely. At (t,α)=(1, 1.8) the 0-indexed sequence 1,1,3,5,10,18,34,… is entirely complete (certified, and directly enumerated gap-free to 300,000). The 1-indexed sequence — same terms minus the leading 1 — is not even cofinitely complete: the integers it fails to represent in [1,N] number 1,290 at N=50,000; 2,007 at 100,000; 3,272 at 200,000; 4,848 at 400,000. The count keeps growing — about 1.2% of all integers permanently missing, scattered forever (a "covered tail" near the top of any finite window is an artifact; the gaps recur above it). One leading ⌊t⌋=1 term fills every one of those ~1.2% of gaps. That is the convention fork, measured.
In the strip 1<α<φ the Lemma-5 cofinite certificate also fires on the 0-indexed sequence, including non-trivially ((t,α)=(5/2,3/2) certifies at r=7 with a subset-sum interval of length 69 ≥ next term 64) — consistent with, and by monotonicity strictly containing, the 1-indexed certificates. The new content is not in the strip but in [φ,2), where Lemma-5 propagation stops working and the greedy-sustainment theorem above carries.
Honest calibration. This is a proof, not a formalization — not Lean/Coq/Isabelle. The analytic part is short enough to read directly (it's reproduced in full above) and every proof-critical computation is exact rational arithmetic, replayable from the linked code. A follow-up round found the closed form: T(α) = min over k≥0 of (2ᵏ+1)/αᵏ is the exact supremum of the entire-complete t's (confirmed to 8·10⁻²⁸ over an adversarial α-set; the powers-of-two greedy prefix 1,2,4,…,2^{k−1} at the boundary is what makes the formula exact). The original wording here overstated the region as the interval [1, t₀(α)] with completeness for all t≤t₀ — that is wrong: entire-completeness is not monotone in t (for α≳1.87 the set can be disconnected, and t=1 stops being entire-complete above α=13^(1/4)≈1.899). Correctly: the entire-complete set contains the top interval [t_min(α), T(α)) and equals [1, T(α)) for α≤3/2 and for 3/2<α≲1.87. Note T(α) bounds entire-completeness, not cofinite-completeness, which is larger and also non-monotone in t (at α=1.618 both conventions are cofinite at t=2.5 > T). The theorem is a sufficient condition, not a characterization: it is silent on cofinite-but-not-entire sequences and on α≥2 (where it correctly breaks down — M→∞ as α→2). The 1-indexed "no complete point for α≥5^(1/3)" is van Doorn's Proposition 3 (verified against the PDF in Round 3); our recurring-gap count above is a consistency check on it, not a re-proof. An adversarial review judged the argument sound within its stated hypotheses; its strongest noted weakness is reach, not correctness.
Open (updated). The closed form for t₀(α) is now in hand (errata above + the follow-up round). Still open: a Lean formalization of the sustainment lemma and the boundary; the exact 0-indexed cofinite frontier in t (a curve, not just spot points — and whether it yields a complete-vs-incomplete separation past t=2 anywhere in [1.71, 2): at (1.7, 2) the 0-indexed sequence shows only ~187 missing integers to 67M but a largest gap near 42M that did not stabilize, so it may be a cofinite island there or not); the t>0 extension; and the full 0-indexed solved map inside [φ,2) — genuinely different from, and larger than, the one the literature has been solving.
Round 5 — the capstone: a closed form, a cross-family proof, and a 1964 structure in disguise. The 0-indexed entirely-complete boundary has a closed form, and it is now triply confirmed: our exact-rational computation (the supremum to 8·10⁻²⁸ over an adversarial α-set), a same-family adversarial pass that caught the non-monotonicity, and a decorrelated GPT-5 Pro proof. The Pro argument is the cleanest of the three and worth naming: its upper bound is a three-line induction — entire-completeness forces a_n ≤ 2^n (since the partial sum C_{n-1} ≤ 2^n − 1), hence t·α^n < 2^n+1 for every n, hence t ≤ min over n of (2^n+1)/α^n, with no monotonicity assumption; its lower bound is a local doubling lemma (⌊αx ≤ 2⌊x⌋ once ⌊x⌋ is large enough) that closes entire-completeness across the whole shelf [(2/α)^k, (2^k+1)/α^k). At the boundary t₀ the next term is 2^k+1 against the ceiling 2^k, so t₀ is a supremum, never attained. (The f(n),M sustainment lemma from Round 4 turns out to be unnecessary — the doubling closes it directly.)
The shape of the complete set. It is not an initial interval. The shelf [(2/α)^k, (2^k+1)/α^k) is always entirely complete (proven), but for many α the entire-complete set extends below the shelf and can even be disconnected: at α=1.9 it is the single interval ≈(1.053, 1.304) (t=1 is not in it — the permanent gap is at 12), while at α=1.95 it is two islands, ≈(1.026, 1.029) and ≈(1.052, 1.170). All of this is machine-confirmed: the boundary prefixes [1,3], [1,2,5], [1,2,4,8,17], [1,2,4,8,16,33] with permanent gaps 2,4,16,32; the shelves proven entire; the α=1.95 island entire while the hole between the islands is not. One clean structural fact falls out of the provenance below: for t<α, complete and entirely-complete coincide (so the cofinite-but-not-entire phenomenon — e.g. (1.618, 2.5), complete with 150 exceptions — can only occur at t≥α).
Where it lives in the literature — the real payoff. Set u = t/α. The 0-indexed sequence at t is, term for term, the literature's sequence at u — the convention fork is just this reparametrization. So for t<α (u<1) we are inside the unit square of Graham, On a conjecture of Erdős in additive number theory, Acta Arithmetica 10 (1964), 63–70, which we read from the primary scan (the official ICM Acta Arithmetica image-PDF, OCR-verified): Lemma 1 (p.64) is our greedy criterion; Lemma 4 (p.65) is the doubling ⌊t·α^{n+1}⌋ ≤ 2⌊t·α^n⌋ once ⌊t·α^n⌋ > (α−1)/(2−α) — i.e. the very move the Pro proof made; Theorem 5(I) (p.68) is the powers-of-two-then-overshoot obstruction, t·α^n < 2^{n-1}+1 and t·α^{n+1} > 2^n+1 (strict), on which our boundary lands exactly; Theorem 6 (p.69) writes the non-complete set as the union over m,n of the regions A_n^(m), B_n^(m), C_n^(m) — so the holes and fragments we found are Graham's regions, pulled back by u=t/α; and Theorem 3 (p.65) is the complete-iff-entirely-complete coincidence in the square (Graham gives the complete set area ≈ 0.85 of the square). For t≥α (u≥1) the classification is van Doorn 2026 (arXiv:2602.23394), whose 1-indexed threshold min(2/α, 3/α², 5/α³) reindexes to our k=0,1,2 piece min(2, 3/α, 5/α²). So the set, its area, and its region structure are Graham's; the 1-indexed closed thresholds are van Doorn's.
So what is genuinely new — honestly. Not the theorem. What is new: (i) the independent cold re-derivation — we conjectured the closed form from exact computation and proved it (re-deriving Graham's doubling along the way) before knowing Graham had it, a genuine reconstruction of a 60-year-old result from a 0-indexed, computation-first angle; (ii) the compact one-line min-formula for the boundary of Graham's set in the 0-indexed parameter — Graham gave the set as the complement of countably many regions, not as a closed form, and the explicit shelf [(2/α)^k, (2^k+1)/α^k) together with the least-minimizer convention k*(α) = max(0, ceil(log2((α−1)/(2−α)))) is, per the literature check, a unification not previously written down; (iii) naming the fragmentation as Graham's regions wearing 0-indexed clothes. That is a real clarity-and-unification contribution plus an independent derivation — creditable, and we believe a better and truer story than claiming a new theorem, which this is not.
Honest calibration. The proof is not Lean-formalized. The primary Graham read was via OCR of an image-only scan: prose and the cited inequalities are clean; a few internal subscripts inside the Theorem 5 proof are garbled and not relied upon; the printed Theorem 5(I) condition 2 is a strict > (not ≥). Van Doorn does not restate Graham's lemmas with matching numbering (his Lemma 1 ≈ Graham's Theorem 1; his Lemma 4 is his own generalization of the doubling), so we cite Graham as the primary source for the lemmas, not via van Doorn. Proven as corollaries: t₀(α)=2 for α≤3/2; t₀(α)→1 as α→2⁻ with the sharp asymptotic t₀(α) = 1 + ((2−α)/(2·log 2))·log(1/(2−α)) + o((2−α)·log(1/(2−α))); and t=1 is entirely complete iff α < 13^(1/4) ≈ 1.899.
Round 5, continued — the cofinite frontier, and the parked question gets a proof. A second Pro pass answered the cofinite question we'd left open, and every load-bearing claim in it machine-checks against our checker. Two indexing identities, both verified: term-for-term A⁰_{t,α} = S¹_{t/α,α} (so the 0-indexed problem is exactly van Doorn's 1-indexed problem with the horizontal parameter rescaled by α, and t₁⁽⁰⁾ = α·t₁⁽¹⁾); and at equal t, A⁰_{t,α} = {⌊t} ⊎ S¹_{t,α} (the identity behind the 4528→150 collapse).
The 4528→150 collapse is exact lag-2 erosion, not vague density. Adding the leading term 2 turns the 1-indexed exception set G into G' = {m≥2 : m∈G and m−2∈G} ∪ {1 if 1∈G} — i.e. R' = R ∪ (2+R). We verified this as a literal set-equality (R⁰ = R | (R≪2), bit-for-bit on [0, 3M]), with |G| = 4528 and |G'| = 150. The mechanism is an exact complementary alignment: every one of the 1793 holes in the tail prefix's critical central band lies two units after an already-representable integer, so the 0-indexed prefix fills them all and Lemma 5 fires at r=25.
The parked (1.7, 2) question is resolved — and the answer is no, it is not a cofinite island. For α≥φ the coefficient 1+α−α² in a_n+a_{n+1}−a_{n+2} ≥ tαⁿ(1+α−α²)−2 turns non-positive, so van Doorn's Lemma 2 gives s_n+s_{n+1} ≤ s_{n+2} for every n and Corollary 1 propagates any seed gap into infinitely many permanent gaps. With ⌊t⌋≥2 the integer 1 is a seed gap (every term is ≥2), so it can never heal: α≥φ and t≥2 ⇒ the 0-indexed sequence is incomplete, with no Prop 3 or 5^(1/3) needed. The "long clean top" our brute force saw at (1.7, 2) was the gap between two of these exponentially-sparse permanent gaps (their ratio tends to α² = 2.9241, confirmed); the formula M_k = 1+Σ_{j=1..k} a_{2j} yields 16 of them ≤50M, all verified genuinely missing, and our observed 42,597,428 is a real permanent gap from a different propagation chain — which only reinforces "infinitely many gaps."
The sub-φ frontier, and how far completeness really reaches. For 1<α<φ the standing conjecture (van Doorn, open) is t₁(α)=∞ — the cofinite region is expected to be all t>0 — so there is no known finite closed form for t₁ there. But it is far larger than t₀: translating our certified shelf gives t₁⁽⁰⁾(α) ≥ 3α for α≤1529/1000, and a concrete new verified point puts t₁⁽⁰⁾(809/500) ≥ 7281/1000 = 7.281 (Lemma-5 certificate reproduced exactly: 1-indexed prefix sum 242,912,917, s₃₆ = 150,120,200, the interval [35,440,664, 207,472,253] of length 172,031,590 > s₃₆) versus t₀(809/500) = 1500/809 ≈ 1.854. Cofinite completeness moves so far beyond entire completeness because it only needs a long interval somewhere in the finite-prefix subset sums; once its length reaches the next term, translated copies overlap and Lemma 5 propagates it forever.
An exact separation, necessarily below t=2. At (α,t) = (13/8, 3/2) the 0-indexed sequence is entirely complete (verified) while the 1-indexed one is incomplete: van Doorn's Proposition 4 gives the 1-indexed threshold b(α) = min(3/α², 5/α³) = 192/169 ≈ 1.136, and 3/2 > b, whereas the 0-indexed reindex 12/13 < b. The t=2 barrier is sharp in mechanism: the moment the leading 0-indexed term jumps from 1 to 2 it creates the seed gap 1, and above φ that gap never heals.
The clean verified picture. Putting it together: for φ ≤ α < 2 the 0-indexed complete set is exactly (0, t₀(α)] — entirely complete on [1, t₀], all of (0,1) complete (since 1/α lies below Graham's first wedge for these α), joined at 1, and nothing at t≥2. For 1 < α < φ it contains (0, t₀(α)] and is conjectured to be all of (0, ∞), with verified points far past t₀ (7.281 at α=1.618). So the convention fork's real content is now fully mapped: a closed-form boundary, a complete-set shape that collapses to a clean interval above φ, and a sub-φ region that bulges without known bound.
Open. A Lean formalization of the doubling lemma, the boundary, and the φ-transition propagation; the sub-φ cofinite frontier t₁(α) (conjectured ∞ — no finite form known, and how far the complete set bulges past t₀ there); and, optionally, a figure overlaying Graham's set Z against our computed complete set. (The (1.7, 2) cofinite-island question is no longer open — see above: incomplete, by the propagated seed gap.)