Status: Open. Attempted by the Committee, August 2026. Not a claimed solution.
Nothing below is offered as a solution of Erdős #348. The official case (m, n) = (2, 3) remains open. What is recorded is a weaker theorem, a negative lemma on arithmetic tails, and a list of constructions that failed the official quantifiers.
1. The problem
A sequence A = {a1 ≤ a2 ≤ ⋯} of positive integers is complete (in the intended, weak sense) if every sufficiently large positive integer is a sum of distinct terms of A. Equivalently: the set P(A) of distinct-term subset sums of A contains all but finitely many positive integers. Finitely many exceptions are allowed. Strong completeness asks that every positive integer, not merely every sufficiently large one, lie in P(A).
On finite prefixes the distinction is visible in the subset-sum bitset. Write σN for the sum of the first N terms. The missing set in {1, …, σN} splits into a finite interior set in {1, …, ⌊σN/2⌋} and traveling high-end mirrors σN − x of those interior misses. The remainder is weakly complete if and only if the interior missing set stays bounded as N → ∞. High-end mirrors are ignored.
T. F. Bloom’s statement of Erdős #348 (Erdős–Graham 1980), quoted from the problem page and its LaTeX source, is as follows.
For what values of 0 ≤ m < n is there a complete sequence A = {a1 ≤ a2 ≤ ⋯} of integers such that
— A remains complete after removing any m elements, but
— A is not complete after removing any n elements?
The word “any” is used in parallel for m and for n. The Lean formalisation writes both quantifiers as universal:
∀ s, |s| = m → IsAddComplete( … ),
∀ t, |t| = n → ¬ IsAddComplete( … ).
Bloom’s “any” is therefore ∀, not ∃. Completeness throughout is the weak one. Strong completeness is already impossible for all 2 ≤ m < n (Brown–Weiss, On N-sequences, Math. Mag. 44 (1971); van Doorn).
Call a complete sequence m-robust if it remains weakly complete after every deletion of m terms, and n-fragile if it fails to be weakly complete after every deletion of n terms. Official #348 asks for an (m, n)-pair: m-robust and n-fragile, both universal. A strictly weaker reading, used below as a search target and not as the problem, is ∃-n-fragile: some (not every) n-deletion leaves unbounded missing sums.
2. Known cases
Two infinite examples are classical, and they fix the meaning of “any.”
Powers of 2 give (m, n) = (0, 1). The sequence 1, 2, 4, 8, … is complete: every positive integer has a unique binary expansion. After any one deletion the missing power of 2 leaves an arithmetic progression of holes. In particular the sequence is not even 1-robust. This is the Brown–Weiss 1-sequence at the weak level.
Fibonacci gives (m, n) = (1, 2), because every 2-deletion fails. Write F1 = F2 = 1 and Fn = Fn−1 + Fn−2, so
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … .
Slack when adding Fn+1 is exactly Fn. The sequence survives deleting any one term (slack 0 in the worst case) and remains strongly complete. After any two deletions, Graham’s argument produces infinitely many missing values of the shape Fs+k+1 − 1. This is not a statement about a particular pair of 1s. Among the first ten Fibonacci terms, all 45 pairs have growing interior miss; early gaps persist at fraction 1. After deleting both 1s, prefix miss counts run 986, 6764, 46367, 317810 at lengths 16, 20, 24, 28. Positive tail slack does not fill the propagating holes. Fibonacci is therefore 1-robust and 2-fragile in the official, universal sense — the same ∀ that Bloom wrote for both m and n.
Graham’s sequence sn = Fn − (−1)n (Fibonacci Quart. 2 (1964)) is weakly complete after any finite deletion and incomplete after any infinite deletion. van Doorn’s sign variant Fn + (−1)n is the same kind of object. Graham is too robust for (2, 3): it cannot fail after three deletions. A (2, 3) witness would have to sit strictly between Fibonacci and Graham.
3. Brown’s criterion
Brown’s criterion is the standard test for strong completeness of a nondecreasing sequence of positive integers: if a1 = 1 and
ak+1 ≤ 1 + ∑i ≤ k ai
for every k, then every positive integer is a subset sum. For nondecreasing positive sequences the inequality is also necessary, and is equivalent to the subset sums of each prefix covering {1, …, ∑ a}. Slack at step k is 1 + σk − ak+1.
After a fixed deletion one cannot read Brown failure at a single index as weak incompleteness. Slack may resume, and a single hole may remain a finite exception. Weak incompleteness requires an unbounded missing set: Fibonacci-style gap propagation, a gcd > 1 obstruction, or a modular gap of Cassels type. Conversely, Brown slack eventually positive is not enough by itself — Fibonacci after two deletions has huge positive tail slack and still infinitely many gaps.
A finite sequence is 2-redundant when every 2-deletion remains Brown-complete, and 3-redundant when every 3-deletion does. Necessity for 2-redundancy: at least three 1s, and for each later term at ≤ 1 + St−1 − M1 − M2, where M1, M2 are the two largest among the prefix of length t − 1. Finite 2-redundant, not 3-redundant sequences exist (the greedy four-one certificate (1,1,1,1,3,4,5,8,12,17,25,37) among others). They do not lift: a finite Brown gap in the tail of an infinite sequence can heal. Moreover there is no finite Brown-complete sequence of length L ≥ 4 for which every 3-deletion is incomplete — the prefix of length L − 3 is Brown-complete, and it is exactly the sequence with the last three terms deleted.
4. Theorem (weaker): three 1s and the even positives
Theorem Let
A = {1, 1, 1} ∪ {2, 4, 6, 8, …} = {1, 1, 1} ∪ 2ℕ,
written in nondecreasing order, the three 1s being the only odd terms. Then A is complete, 2-robust, and ∃-3-fragile: after deleting the three 1s, every subset sum is even, so the missing set is unbounded.
This is not official #348. The 3-fragility is existential, not universal. The argument occupies the rest of this section.
Completeness of A
Undeleted prefixes have interior miss 0. Brown holds: the next even is 2k + 2 ≤ 1 + 3 + k(k + 1).
∃-3-fragility
Proof Delete the three 1s. The remainder is {2, 4, 6, …}. Every subset sum is even. Every odd positive integer is missing. At prefix length N the interior miss is exactly σ/2 odds, growing, with persistence 1. Density 1/2. □
Prefix counts after this deletion:
2-robustness
Proof After any 2-deletion at least one 1 remains, so gcd = 1. Three cases.
Two 1s deleted. The remainder is {1} ∪ {2, 4, 6, …}. The evens give all even positives; adding the remaining 1 gives all odds. The remainder is strongly complete. Prefix miss is 0 at N = 16, 24, 32, 40.
One 1 and one even 2k deleted.
If k = 1 (delete 1 and 2): the remainder is {1, 1, 4, 6, 8, …}. It makes 1 and 2; it misses only 3; it makes all integers ≥ 4. The interior miss is {3}, frozen. Prefix miss is 2 (the 3 and its σ-mirror) at every N = 16..40.
If k = 2 (delete 1 and 4): the remainder misses only 5. Frozen.
If k ≥ 3: the prefix {1, 1} ∪ {2, 4, …, 2k − 2} covers 1 through its sum 2 + k(k − 1). The next remaining even is 2k + 2 ≤ k2 − k + 3 for k ≥ 3. Brown continues. The remainder is strongly complete.
Two evens 2j < 2k deleted. Three 1s remain.
If (2, 4): the remainder is {1, 1, 1, 6, 8, …}. It makes 1, 2, 3; it misses 4, 5; it makes all integers ≥ 6. Frozen.
If (4, 6): it misses 6, 7. Frozen. (Brown fails at 8 > 1+1+1+2 = 5, but only two holes.)
If (2, 6), and all k ≥ j + 2, and consecutive pairs with j ≥ 3: the prefix of three 1s plus the remaining small evens covers a run long enough that the next even satisfies Brown. Strongly complete.
Thus every 2-deletion is weakly complete. The only interior holes that ever appear are among {3}, {5}, {4, 5}, {6, 7}.
Exhaustive 2-deletions of prefixes confirm the list is closed: the number of pairs with miss > 0 is exactly 8 at every N ∈ {16, 20, 24, 28} (three choices of which 1 pairs with 2, three with 4, plus (2, 4) and (4, 6)). No new 2-failure appears as the even tail lengthens. All other pairs among the first N terms, save those eight, have miss = 0.
The odd-count threshold
Three odds is exact.
5. Why this is not official #348
Official #348 wants every 3-deletion to destroy weak completeness. The sequence A fails that. Only 3-deletions that remove all three odds produce unbounded misses. The others heal.
Delete (2, 4, 6): the interior miss is {4, 5, 6, 7}, frozen at miss = 8 (mirrors included) through N = 32. Delete three large consecutive evens: miss = 0. Delete two 1s and one late even: miss = 0.
This is not an accident of the even tail. It is forced by any arithmetic progression as a tail.
Lemma (AP-tail) Let G be a finite multiset of positive integers, M ≥ 1, and T = {M, 2M, 3M, …}. If A = G ∪ T is weakly complete, then for every finite F ⊂ T, the remainder A ∖ F is still weakly complete.
Proof Subset sums of {1, 2, 3, …} cover every positive integer. Deleting finitely many elements from that sequence leaves it weakly complete, so P(T ∖ F) contains every sufficiently large multiple of M. Weak completeness of A forces P(G) to hit every residue class modulo M. Hence P(G) + P(T ∖ F) contains every sufficiently large integer. □
Corollary No sequence of the form G ∪ Mℕ with G finite is 3-deletion-universal-fragile: every 3-deletion contained in the tail heals. The only 3-deletions that can fail are those that strip enough of G to lose a residue class modulo M, and there are only finitely many such triples.
Adding a second modulus enlarges the finite list of modular breaking triples and does not produce ∀-3. For
A = {1, 1, 1, 2, 2, 2} ∪ 6ℕ,
residues modulo 6 give eleven modular breaking triples (the three 1s; the three 2s; nine of type (1, 2, 2)) and no others. Exhaustive 220 triples among the first 12 terms, interior miss tracked to N = 36: 11 grow with persistence 1, matching the residue table; 203 heal. Two-mod is 2-robust (any 2-deletion leaves a small remainder that still hits every residue modulo 6; the tail is 6ℕ minus at most two terms). It is a strictly larger ∃-3 object than A, and still not ∀-3.
The official (2, 3) case is therefore untouched. ∀-3 remains open.
6. The density trap
To break tail 3-deletions one needs a non-arithmetic (typically exponential) tail. Making that tail sparse enough for 3-deletions to propagate gaps typically makes some 2-deletions propagate as well. Extra density that stops 2-deletion gap growth also stops 3-deletion gap growth — unless the 3-failure is modular, or the sequence is strictly faster than tight 2-Brown by one unit. The modular exception is the gcd construction above, and it cannot be ∀-3 by the AP-tail lemma. The one-unit exception is computational and still has healing triples.
Qualitatively, along linear recurrences:
- Ratio φ (Fibonacci): 1-robust, 2-fragile via gap propagation, even though tail Brown slack tends to infinity.
- Ratio < φ (Narayana, Padovan, an = an−1 + an−4): 2-robust, and 3-deletions leave only finite holes — slack absorbs any fixed deletion and extra representations fill gaps.
- Occasional φ-sized jumps: some 2-deletions start propagating gaps again.
- Multiplicity on Fibonacci: 2-robust, and most 3-deletions heal; fragility returns after deleting both copies of a doubled value together with the successor, and not after a generic triple.
Tight 2-Brown from three 1s is Narayana’s cows, an = an−1 + an−3,
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, … ,
ratio the supergolden root of x3 = x2 + 1 ≈ 1.465 < φ. Identity σn = an+3 − 1; slack when adding an+1 is exactly an + an−1. Every tested 2-deletion has miss = 0 (all 91 pairs among the first 14 indices on a length-24 prefix). Every tested 3-deletion has frozen finite miss, including mid-index triples with a long tail. Narayana is a strong 2-sequence that fails 3-fragility even in the weak, existential sense.
Shifting the 2-Brown recurrence by one unit of extra,
an+1 = max(an, 1 + σn − M1 − M2 + extra),
from three 1s: extra = 0 is Narayana; extra = −1 is denser and 3-robust on the tested triples; extra = +2 is not 2-robust; extra = +1 is the greedy sequence 1, 1, 1, 3, 4, 5, 8, 12, 17, … with tail identity an = an−1 + an−4 after the skip of 2. After deleting the three 1s, interior miss grows with persistence 1 through σ ≈ 1.6·106 (not a modular obstruction: gcd remains 1). Scattered triples heal (miss = 0). This is a computational weaker-(2, 3) witness, not a closed-form proof that the missing set is infinite, and not official #348.
The same trap appears when one thins 6ℕ to beat the AP-tail lemma. Greedy 2-Brown rounded up to a multiple of 6, and the geometric tail {1,1,1,2,2,2} ∪ {6·2k}, both make some consecutive tail triples grow — and both make 2-deletions of the tail grow. Density that protects 2-deletions fills 3-deletions of the tail; sparsity that breaks tail 3-deletions breaks some 2-deletions.
7. What was tried and failed
No construction below is a solution of official #348. Several are intermediate objects: 2-robust, with some (even infinitely many) breaking 3-triples, and a majority of healing triples.
8. What remains
Official Erdős #348, case (m, n) = (2, 3), is open for weak completeness. Strong (2, 3) is already impossible.
A genuine official (2, 3) would need a representation system in which every 3-set seeds unbounded gap propagation while no 2-set does. Obstacles that did not move:
- Gcd constructions kill only the triples that strip every odd (or every residue coprime to d). All other triples heal. The AP-tail lemma rules out every finite generator plus an arithmetic tail.
- extra = +1 and the forum Fibonacci-multiplicity sequence produce infinitely many breaking consecutive triples with propagating interior holes, but scattered triples have miss = 0. The gap between “infinitely many breaking triples” and “all triples break” is the whole difficulty.
- Finite prefixes can never certify “every 3-deletion”: deleting the last three terms of a prefix is not a statement about the infinite sequence, and any Brown-complete prefix of length ≥ 3 is complete.
- The density trap. Density that protects every 2-deletion fills 3-deletions of the tail; sparsity that breaks every tail 3-deletion breaks some 2-deletion. Linear recurrences with growing slack cannot keep a fixed 3-deletion incomplete forever without opening a 2-deletion at those same tight indices.
The only object in this note with a complete proof is the weaker sequence A = {1, 1, 1} ∪ 2ℕ: complete, 2-robust, and ∃-3-fragile. It is not a witness for Bloom’s problem. Nothing here is claimed as a solution.
T. F. Bloom, Erdős Problem #348, https://www.erdosproblems.com/348. LaTeX source: https://www.erdosproblems.com/latex/348. Formalisation: FormalConjectures/ErdosProblems/348.lean. Brown–Weiss, On N-sequences, Math. Mag. 44 (1971). Graham, Fibonacci Quart. 2 (1964). van Doorn on strong (m, n).
