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Among others, we prove that $$ \\sum_{k=0}^{[n/4]}(-1)^k {n\\choose 4k}\\frac{B_{n-4k}(z) }{2^{6k}} =\\frac{1}{2^{n+1}}\\sum_{k=0}^{n} (-1)^k \\frac{1+i^k}{(1+i)^k} {n\\choose k}{B_{n-k}(2z)} $$ and $$ \\sum_{k=1}^{n} 2^{2k-1} {2n\\choose 2k-1} B_{2k-1}(z) =\n\\sum_{k=1}^n k \\, 2^{2k} {2n\\choose 2k} E_{2k-1}(z). $$ Applications of our results lead to formulas for Bernoulli and Euler numbers, like, for instance, $$ n E_{n-1} =\\sum_{k=1}^{[n/2]} \\frac{2^{2k}-1}{k} (2^{2k}-2^n){n\\choose 2k-1} B_{2k}B_{n-2k}. $$"},"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":"1710.07127","kind":"arxiv","version":1},"metadata":{"license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","primary_cat":"math.CA","submitted_at":"2017-10-19T13:18:52Z","cross_cats_sorted":[],"title_canon_sha256":"735d3bb4f9aa7feaec2bbd6a96bf9193b8d7992990f8644bc273ed9d809c2f58","abstract_canon_sha256":"78cc170a7fde2a6066e5846317f1739176ce6ba1a44af11d0217b86f764efd05"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-18T00:32:26.928102Z","signature_b64":"Bckx0TUwCSyyDRfFJiiOFdmBHmneQFIdExjuSusyORJHVmx69vwtGpOyHB2x7oFKstWCLifD7L8o4LBRKdQmDg==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"1dc95ed2d817060d396c19038ef354575857b193c0ad13ca552f310bfb770b29","last_reissued_at":"2026-05-18T00:32:26.927449Z","signature_status":"signed_v1","first_computed_at":"2026-05-18T00:32:26.927449Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Identities involving Bernoulli and Euler polynomials","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.CA","authors_text":"Horst Alzer, Semyon Yakubovich","submitted_at":"2017-10-19T13:18:52Z","abstract_excerpt":"We present various identities involving the classical Bernoulli and Euler polynomials. Among others, we prove that $$ \\sum_{k=0}^{[n/4]}(-1)^k {n\\choose 4k}\\frac{B_{n-4k}(z) }{2^{6k}} =\\frac{1}{2^{n+1}}\\sum_{k=0}^{n} (-1)^k \\frac{1+i^k}{(1+i)^k} {n\\choose k}{B_{n-k}(2z)} $$ and $$ \\sum_{k=1}^{n} 2^{2k-1} {2n\\choose 2k-1} B_{2k-1}(z) =\n\\sum_{k=1}^n k \\, 2^{2k} {2n\\choose 2k} E_{2k-1}(z). $$ Applications of our results lead to formulas for Bernoulli and Euler numbers, like, for instance, $$ n E_{n-1} =\\sum_{k=1}^{[n/2]} \\frac{2^{2k}-1}{k} (2^{2k}-2^n){n\\choose 2k-1} B_{2k}B_{n-2k}. $$"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1710.07127","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"},"aliases":[{"alias_kind":"arxiv","alias_value":"1710.07127","created_at":"2026-05-18T00:32:26.927543+00:00"},{"alias_kind":"arxiv_version","alias_value":"1710.07127v1","created_at":"2026-05-18T00:32:26.927543+00:00"},{"alias_kind":"doi","alias_value":"10.48550/arxiv.1710.07127","created_at":"2026-05-18T00:32:26.927543+00:00"},{"alias_kind":"pith_short_12","alias_value":"DXEV5UWYC4DA","created_at":"2026-05-18T12:31:12.930513+00:00"},{"alias_kind":"pith_short_16","alias_value":"DXEV5UWYC4DA2OLM","created_at":"2026-05-18T12:31:12.930513+00:00"},{"alias_kind":"pith_short_8","alias_value":"DXEV5UWY","created_at":"2026-05-18T12:31:12.930513+00:00"}],"events":[],"event_summary":{},"paper_claims":[],"inbound_citations":{"count":0,"internal_anchor_count":0,"sample":[]},"formal_canon":{"evidence_count":0,"sample":[],"anchors":[]},"links":{"html":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5","json":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5.json","graph_json":"https://pith.science/api/pith-number/DXEV5UWYC4DA2OLMDEBY542UK5/graph.json","events_json":"https://pith.science/api/pith-number/DXEV5UWYC4DA2OLMDEBY542UK5/events.json","paper":"https://pith.science/paper/DXEV5UWY"},"agent_actions":{"view_html":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5","download_json":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5.json","view_paper":"https://pith.science/paper/DXEV5UWY","resolve_alias":"https://pith.science/api/pith-number/resolve?arxiv=1710.07127&json=true","fetch_graph":"https://pith.science/api/pith-number/DXEV5UWYC4DA2OLMDEBY542UK5/graph.json","fetch_events":"https://pith.science/api/pith-number/DXEV5UWYC4DA2OLMDEBY542UK5/events.json","actions":{"anchor_timestamp":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5/action/timestamp_anchor","attest_storage":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5/action/storage_attestation","attest_author":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5/action/author_attestation","sign_citation":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5/action/citation_signature","submit_replication":"https://pith.science/pith/DXEV5UWYC4DA2OLMDEBY542UK5/action/replication_record"}},"created_at":"2026-05-18T00:32:26.927543+00:00","updated_at":"2026-05-18T00:32:26.927543+00:00"}