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We give three short proofs. The first is two lines: subtract the defining recurrence at adjacent indices and the constant cancels (we call this homogenisation). The second reads off the same relation from the exponential generating function $F(x) = "},"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":true,"formal_links_present":false},"canonical_record":{"source":{"id":"2605.15500","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CO","submitted_at":"2026-05-15T00:36:49Z","cross_cats_sorted":[],"title_canon_sha256":"e0d9c94d9e3dc39e18cd2e415b03d91da827af5d74021cf6d6d7ab95cd1d0950","abstract_canon_sha256":"e2764533384f774269f767e12f34025b5d287aeb7b89a8c5e29250c2e5f24110"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-20T00:01:01.861138Z","signature_b64":"lSigArlTC5au/KgkfYaf3QiAstZyB63IlrD9+Cr0CKHMY2NnGF8asBBUMPMtDunng96Z507h9U9QSQPHeg4JDg==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"1a82742dbd98ebdf0ef8844109d34715fce10c2c8f811c4dc3a0e9ba4aa2d8cb","last_reissued_at":"2026-05-20T00:01:01.860347Z","signature_status":"signed_v1","first_computed_at":"2026-05-20T00:01:01.860347Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Three short proofs of Mathar's 2014 conjecture for OEIS A002627","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"The sequence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2.","cross_cats":[],"primary_cat":"math.CO","authors_text":"Tong Niu","submitted_at":"2026-05-15T00:36:49Z","abstract_excerpt":"For the OEIS sequence A002627, defined by the inhomogeneous first-order recurrence $a(n) = n\\,a(n-1) + 1$ with $a(0) = 0$, R.~J.~Mathar recorded in February 2014 the conjectured second-order homogeneous recurrence \\[ a(n) - (n+1)\\,a(n-1) + (n-1)\\,a(n-2) = 0, \\qquad n \\ge 2, \\] which has remained marked as a conjecture on the OEIS for over a decade. We give three short proofs. The first is two lines: subtract the defining recurrence at adjacent indices and the constant cancels (we call this homogenisation). The second reads off the same relation from the exponential generating function $F(x) = "},"claims":{"count":4,"items":[{"kind":"strongest_claim","text":"The sequence a(n) defined by a(n) = n a(n-1) + 1, a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2.","source":"verdict.strongest_claim","status":"machine_extracted","claim_id":"C1","attestation":"unclaimed"},{"kind":"weakest_assumption","text":"The first-order recurrence a(k) = k a(k-1) + 1 holds exactly for every integer k >= 1, so that subtraction at adjacent indices is valid and produces no boundary or remainder terms.","source":"verdict.weakest_assumption","status":"machine_extracted","claim_id":"C2","attestation":"unclaimed"},{"kind":"one_line_summary","text":"Three short elementary proofs establish the conjectured homogeneous recurrence for OEIS A002627 from its defining inhomogeneous recurrence.","source":"verdict.one_line_summary","status":"machine_extracted","claim_id":"C3","attestation":"unclaimed"},{"kind":"headline","text":"The sequence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2.","source":"verdict.pith_extraction.headline","status":"machine_extracted","claim_id":"C4","attestation":"unclaimed"}],"snapshot_sha256":"8387d31337b9ed5809dee261dc9f2d69eaa439a4af9d84d1d2eb12766678aecb"},"source":{"id":"2605.15500","kind":"arxiv","version":1},"verdict":{"id":"43171870-8907-4fc0-bc9f-3e34b8dbd2ad","model_set":{"reader":"grok-4.3"},"created_at":"2026-05-19T15:47:43.264907Z","strongest_claim":"The sequence a(n) defined by a(n) = n a(n-1) + 1, a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2.","one_line_summary":"Three short elementary proofs establish the conjectured homogeneous recurrence for OEIS A002627 from its defining inhomogeneous recurrence.","pipeline_version":"pith-pipeline@v0.9.0","weakest_assumption":"The first-order recurrence a(k) = k a(k-1) + 1 holds exactly for every integer k >= 1, so that subtraction at adjacent indices is valid and produces no boundary or remainder terms.","pith_extraction_headline":"The sequence a(n) = n a(n-1) + 1 with a(0) = 0 satisfies a(n) - (n+1) a(n-1) + (n-1) a(n-2) = 0 for all n >= 2."},"integrity":{"clean":true,"summary":{"advisory":0,"critical":0,"by_detector":{},"informational":0},"endpoint":"/pith/2605.15500/integrity.json","findings":[],"available":true,"detectors_run":[{"name":"doi_title_agreement","ran_at":"2026-05-19T16:01:17.959288Z","status":"completed","version":"1.0.0","findings_count":0},{"name":"doi_compliance","ran_at":"2026-05-19T15:53:35.331222Z","status":"completed","version":"1.0.0","findings_count":0},{"name":"cited_work_retraction","ran_at":"2026-05-19T14:51:56.089870Z","status":"completed","version":"1.0.0","findings_count":0},{"name":"claim_evidence","ran_at":"2026-05-19T14:21:54.064323Z","status":"completed","version":"1.0.0","findings_count":0},{"name":"shingle_duplication","ran_at":"2026-05-19T13:49:41.855084Z","status":"skipped","version":"0.1.0","findings_count":0},{"name":"citation_quote_validity","ran_at":"2026-05-19T13:49:41.394160Z","status":"skipped","version":"0.1.0","findings_count":0},{"name":"ai_meta_artifact","ran_at":"2026-05-19T13:33:22.642164Z","status":"skipped","version":"1.0.0","findings_count":0}],"snapshot_sha256":"c19c1cf6e79b7189752a7ca328211caae70061783b30ed00729d62a4fa861220"},"references":{"count":11,"sample":[{"doi":"","year":2023,"title":"M. 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