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We also give nearly matching upper and lower bounds for large $n$, showing as a corollary that $f(n) \\approx n/2$."},"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":"1807.10231","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CO","submitted_at":"2018-07-26T16:33:50Z","cross_cats_sorted":["math.AT"],"title_canon_sha256":"1056b2b61a23ee23b211347569ac8e3b0fa59a775417953342c148ae957c9257","abstract_canon_sha256":"b9a7a98a526f75abc2d9e7c3c84f8af55bf16e3bc5b45c076c7cfa828e485513"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:09:42.787797Z","signature_b64":"+VmANQADU8pxB6ZTZPakDZ47YvwGBY5jHrXpwhRSBx7Cqdf3O8wdM7I2Yq7gaHbbc7gwcOPO0192BjB4Q1XnBQ==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"b181cf42f34c173a1ef58d774981b6aeda4c35245ad407f07367bbee240072cf","last_reissued_at":"2026-05-18T00:09:42.787246Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:09:42.787246Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Polyominoes with maximally many holes","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.AT"],"primary_cat":"math.CO","authors_text":"\\'Erika Rold\\'an, Matthew Kahle","submitted_at":"2018-07-26T16:33:50Z","abstract_excerpt":"What is the maximum number of holes that a polyomino with $n$ tiles can enclose? Call this number $f(n)$. We show that if $n_k = \\left( 2^{2k+1} + 3 \\cdot 2^{k+1}+4 \\right) / 3$ and $h_k = \\left( 2^{2k}-1 \\right) /3$, then $f(n_k) = h_k$ for $k \\ge 1$. 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