Nathanson’s Additive h-Basis Strict-Inequality Refutation

Abstract

The strict inequality in Nathanson’s Problem 12(2) fails for a three-element integer set.

Here h • A denotes the h-fold sumset A + … + A, with repetitions allowed. val denotes the coercion from natural numbers to integers. The expression sSup is the natural-number supremum. On the source domain used by the claim, the relevant sets are nonempty and bounded, so this supremum is a maximum.

Problem 12. It is natural to ask how the use of negative numbers changes the size of the maximal interval [0, n] contained in an h-fold sumset. For integers h ≥ 2 and k ≥ 2, are the following statements true or false? (1) If A ∈ (ℤ choose k) with min(A) < 0, then ℓ_h(A) ≤ n♭_h(k). (2) If A ∈ (ℤ choose k) with min(A) < 0, then ℓ_h(A) < n♭_h(k). Only statement (2) is settled here.

Definition 1.1 (Largest covered initial segment).

$$\forall h\in \mathbb{N},\forall A\in \mathrm{Finset}\left(\mathbb{Z}\right),\mathrm{segmentLength}\left(h,A\right)=\mathrm{sSup}\left(\{n\in \mathbb{N}\mid \forall i\in \mathbb{N},(i\le n)\Rightarrow (\mathrm{val}\left(i\right)\in \mathrm{nsmul}\left(h,A\right))\}\right)$$

Formalization. D5/S3/Arith/NathansonAdditiveHBasisRefutation.segmentLength (✓ std3).

Citation. Melvyn B. Nathanson (2026). Problems in additive number theory, VII: The structure of additive h-bases for n. URL: https://arxiv.org/abs/2605.26425v3.

Commentary.

For h in N and a finite integer set A, segmentLength(h,A) is the largest natural n for which every natural i <= n, coerced to an integer, belongs to h • A.

Definition 1.2 (Maximum over nonnegative k-sets).

$$\forall h\in \mathbb{N},\forall k\in \mathbb{N},\mathrm{nonnegativeMaximum}\left(h,k\right)=\mathrm{sSup}\left(\{n\in \mathbb{N}\mid \exists B\in \mathrm{Finset}\left(\mathbb{N}\right),(\mathrm{card}\left(B\right)=k)\wedge (\forall i\in \mathbb{N},(i\le n)\Rightarrow (i\in \mathrm{nsmul}\left(h,B\right)))\}\right)$$

Formalization. D5/S3/Arith/NathansonAdditiveHBasisRefutation.nonnegativeMaximum (✓ std3).

Citation. Melvyn B. Nathanson (2026). Problems in additive number theory, VII: The structure of additive h-bases for n. URL: https://arxiv.org/abs/2605.26425v3.

Commentary.

For h,k in N, nonnegativeMaximum(h,k) is the largest covered endpoint among all k-element finite subsets B of N. This is n♭_h(k).

Definition 1.3 (The strict universal claim).

$$(claim)\iff (\forall h\in \mathbb{N},\forall k\in \mathbb{N},(2\le h)\Rightarrow \left((2\le k)\Rightarrow (\forall A\in \mathrm{Finset}\left(\mathbb{Z}\right),(\mathrm{card}\left(A\right)=k)\Rightarrow \left((\exists a\in \mathbb{Z},(a\in A)\wedge (a<0))\Rightarrow \left((0\in \mathrm{nsmul}\left(h,A\right))\Rightarrow (\mathrm{segmentLength}\left(h,A\right)<\mathrm{nonnegativeMaximum}\left(h,k\right))\right)\right))\right))$$

Formalization. D5/S3/Arith/NathansonAdditiveHBasisRefutation.claim (✓ std3).

Citation. Melvyn B. Nathanson (2026). Problems in additive number theory, VII: The structure of additive h-bases for n. URL: https://arxiv.org/abs/2605.26425v3.

Commentary.

The claim quantifies h >= 2, k >= 2, and every k-element finite integer set A that contains a negative element and satisfies 0 in h • A. The last premise excludes exactly the cases in which the source leaves ℓ_h(A) undefined.

Theorem 1.4 (Equality at a negative three-set).

$$\neg claim$$

Proof. Machine-checked in Lean as D5/S3/Arith/NathansonAdditiveHBasisRefutation.result (✓ std3). ∎

Resolves. Problems/nathanson-additive-h-bases-problem-12 (refuted) by D5/S3/Arith/NathansonAdditiveHBasisRefutation.result.

Source. Repository-derived.

Acknowledgement. Melvyn B. Nathanson (2026). Problems in additive number theory, VII: The structure of additive h-bases for n. URL: https://arxiv.org/abs/2605.26425v3.

Commentary.

Take h=2, k=3, and A={-1,1,2}. Then 2A={-2,0,1,2,3,4}, so ℓ_2(A)=4. The set {0,1,2} shows n♭_2(3)>=4. Conversely, a three-element B subset N whose double sumset covers 0 through 5 must contain 0 and 1. Representing 3 forces its third element to be 2 or 3, but neither {0,1,2} nor {0,1,3} represents 5. Hence n♭_2(3)=4, contradicting the proposed strict inequality.

References

• Truth anchor: D5/S3/Arith/NathansonAdditiveHBasisRefutation.claim
• Truth anchor: D5/S3/Arith/NathansonAdditiveHBasisRefutation.nonnegativeMaximum
• Truth anchor: D5/S3/Arith/NathansonAdditiveHBasisRefutation.result
• Truth anchor: D5/S3/Arith/NathansonAdditiveHBasisRefutation.segmentLength