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Then for any integer $n$ such that \\[n \\ge \\max\\left\\{k,\\frac{1}{2(t+2)}k^2 + \\frac{q-s}{t+2}k - \\frac{t}{2} + s\\right\\}\\] and any function $f:[n]\\to \\{-1,1\\}$ with $|\\sum_{i=1}^nf(i)| \\le q$, there is a set $B \\subseteq [n]$ of $k$ consecutive integers with $|\\sum_{y\\in B}f(y)| \\le t$. Moreover, this bound is sharp for all the pa"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"1612.06523","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CO","submitted_at":"2016-12-20T06:35:53Z","cross_cats_sorted":["math.NT"],"title_canon_sha256":"ae28686b7c04d449b1348ca2da4b1192fb35ae4264d4a44b184d9b1ca4a392b4","abstract_canon_sha256":"d627dbb5dace9182c7dd74cdecb41b940d42805b7e2f2553509fd65a968add01"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:54:25.534674Z","signature_b64":"XkDJPGd69+zvokZaQ9Mwu97f/Y0eZmmSB609EbsyTQ39Md6uH2u0h6Eh5QC0tqC2epL1YefvjvZE7dkA0/OqAA==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"1107774a0d8d76ce515d94373b05a82314203ad385d9e12f30f4ebab17105984","last_reissued_at":"2026-05-18T00:54:25.533990Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:54:25.533990Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Zero-sum subsequences in bounded-sum $\\{-1, 1\\}$-sequences","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.NT"],"primary_cat":"math.CO","authors_text":"Adriana Hansberg, Amanda Montejano, Yair Caro","submitted_at":"2016-12-20T06:35:53Z","abstract_excerpt":"The following result gives the flavor of this paper: Let $t$, $k$ and $q$ be integers such that $q\\geq 0$, $0\\leq t < k$ and $t \\equiv k \\,({\\rm mod}\\, 2)$, and let $s\\in [0,t+1]$ be the unique integer satisfying $s \\equiv q + \\frac{k-t-2}{2} \\,({\\rm mod} \\, (t+2))$. Then for any integer $n$ such that \\[n \\ge \\max\\left\\{k,\\frac{1}{2(t+2)}k^2 + \\frac{q-s}{t+2}k - \\frac{t}{2} + s\\right\\}\\] and any function $f:[n]\\to \\{-1,1\\}$ with $|\\sum_{i=1}^nf(i)| \\le q$, there is a set $B \\subseteq [n]$ of $k$ consecutive integers with $|\\sum_{y\\in B}f(y)| \\le t$. 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