Problem packetResearch packetR65
Legendre factorization and divisor-count replay
Link to a section
Executable material is recorded. Successful replay is a separate check.
Recorded status: available
Recorded scope: the complete prime-exponent factorizations at (19971,9949) and (1000000,499985)
Complete recorded scope and conditions
{
"kind": "bounded",
"statement": "the complete prime-exponent factorizations at (19971,9949) and (1000000,499985)",
"bounds": {
"pairs": {
"min": 2,
"max": 2
},
"largest_n": {
"min": 1000000,
"max": 1000000
}
},
"exhaustive": true
}Originating problem: Most divisors of a binomial coefficient with top at most 10^6
Recorded relationships: The exact record through n=20,000 occurs at (19,971, 9,949)
Other recorded relationships (1)
Authored record and scope
- Authored title
- Legendre factorization and divisor-count replay
- Record type
- artifact
- Stored status
- available
- Evidence grade
- executable
- Recorded scope data
- { "kind": "bounded", "statement": "the complete prime-exponent factorizations at (19971,9949) and (1000000,499985)", "bounds": { "pairs": { "min": 2, "max": 2 }, "largest_n": { "min": 1000000, "max": 1000000 } }, "exhaustive": true }
- Linked research record IDs
- R68 R67
2Authored explanation
The verifier sieves every prime through each top parameter. Legendre's factorial valuation gives every binomial exponent as an integer. The code joins the nonzero pairs in ascending prime order using `p^e` and hashes the resulting factorization. It then multiplies every \(e+1\), hashes the exact decimal divisor count, and reports each exponent histogram.
The prefix factorization contains 1,542 terms. The cutoff-scale factorization contains 53,478 terms. Their report payload has SHA-256 digest `fea7d94ea1b9b59e253076f13b8caf82c47db0d875c7e0b71d351554ee7eb723`.
Files and source
Files embedded in this record. Matching a file hash confirms its identity.
- R65.txt2,055 bytes · No SHA-256 recorded
Preview R65.txt
from hashlib import sha256 from json import dumps from math import isqrt CASES = [ (19971, 9949, 1542, "11d4dc8a9b2dc54225e841546f236a3a9cd5c2f61fe0db6ab2455f28749b9c6c", "87551dd1380ee16fd2443feb0ea48da980013632cae016af476f8981a8f14683"), (1000000, 499985, 53478, "fb3db7328c0c940f68d88e873c6554c9ec65616c825f5b933fd2e55e492e1be2", "37e6b0aec8c146fa82e6e8d0eb776dbb1504fb2fff80e5fa74bff8eaddfee951"), ] def primes_through(limit): sieve = bytearray(b"\x01") * (limit + 1) sieve[:2] = b"\x00\x00" for p in range(2, isqrt(limit) + 1): if sieve[p]: sieve[p*p:limit+1:p] = b"\x00" * ((limit-p*p)//p+1) return [p for p in range(2, limit + 1) if sieve[p]] def factorial_valuation(n, p): value = 0 while n: n //= p value += n return value def certify(n, k, expected_terms, expected_factor_sha, expected_tau_sha): factors = [] tau = 1 histogram = {} for p in primes_through(n): exponent = factorial_valuation(n,p)-factorial_valuation(k,p)-factorial_valuation(n-k,p) assert exponent >= 0 if exponent: factors.append((p,exponent)) tau *= exponent+1 histogram[exponent] = histogram.get(exponent,0)+1 factor_data = "*".join(f"{p}^{e}" for p,e in factors) tau_text = str(tau) assert len(factors) == expected_terms assert sha256(factor_data.encode()).hexdigest() == expected_factor_sha assert sha256(tau_text.encode()).hexdigest() == expected_tau_sha return { "n":n, "k":k, "factor_terms":len(factors), "exponent_histogram":sorted(histogram.items()), "factor_sha256":expected_factor_sha, "tau_digits":len(tau_text), "tau_prefix":tau_text[:32], "tau_suffix":tau_text[-32:], "tau_sha256":expected_tau_sha, } report = [certify(*case) for case in CASES] payload = dumps(report,sort_keys=True,separators=(",",":")) assert sha256(payload.encode()).hexdigest() == "fea7d94ea1b9b59e253076f13b8caf82c47db0d875c7e0b71d351554ee7eb723" print(payload)File identity
- Recorded filename
- R65.txt
- Download SHA-256
- 671decfbefffc2c29af13bba506d59b87cb6d797038a0b1664904a0361bbc563
Continue this work
Replay material: partial
4Reproduce
Part of the replay path is recorded. Check the missing fields before comparing a new run.
Verification source: doi.org ↗, Inline CPython standard-library computation reproduced on 2026-07-25
Missing for a complete replay: command, expected output.
Recorded artifact fields
5What it produced
6How it connects
Evidence for
- claim
- claim
Recorded for
- problem
Cite this record
Cite the original sources separately.
Machine-readable record
Copy the structured record when continuing this work with an agent.
{
"schema": "theoremdb-agent-record-v1",
"ref": "R65",
"content_hash": null,
"slug": "bdr1m-artifact-factorization-replay",
"type": "artifact",
"title": "Legendre factorization and divisor-count replay",
"summary": "Standard-library Python regenerates both full factorizations and checks their SHA-256 digests.",
"relevance": "For Most divisors of a binomial coefficient with top at most 10^6, record bdr1m-artifact-factorization-replay (“Legendre factorization and divisor-count replay”) supplies evidence or a replay used to check the packet. The record states: Standard-library Python regenerates both full factorizations and checks their SHA-256 digests.",
"relevance_source": "recorded",
"body": "The verifier sieves every prime through each top parameter. Legendre's factorial valuation gives every binomial exponent as an integer. The code joins the nonzero pairs in ascending prime order using `p^e` and hashes the resulting factorization. It then multiplies every \\(e+1\\), hashes the exact decimal divisor count, and reports each exponent histogram.\n\nThe prefix factorization contains 1,542 terms. The cutoff-scale factorization contains 53,478 terms. Their report payload has SHA-256 digest `fea7d94ea1b9b59e253076f13b8caf82c47db0d875c7e0b71d351554ee7eb723`.",
"status": "available",
"evidence_grade": "executable",
"scope": {
"kind": "bounded",
"statement": "the complete prime-exponent factorizations at (19971,9949) and (1000000,499985)",
"bounds": {
"pairs": {
"min": 2,
"max": 2
},
"largest_n": {
"min": 1000000,
"max": 1000000
}
},
"exhaustive": true
},
"reproduction": {
"schema": "theoremdb-reproduction-v1",
"readiness": "partial",
"kind": "inline_python_computation",
"entrypoint": "join source_lines with newline and run with python3",
"runtime": "CPython 3.9 or later, standard library only",
"citation": {
"url": "https://doi.org/10.1134/S0001434613010331",
"locator": "Inline CPython standard-library computation reproduced on 2026-07-25"
},
"inline_source": [
"from hashlib import sha256",
"from json import dumps",
"from math import isqrt",
"",
"CASES = [",
" (19971, 9949, 1542, \"11d4dc8a9b2dc54225e841546f236a3a9cd5c2f61fe0db6ab2455f28749b9c6c\",",
" \"87551dd1380ee16fd2443feb0ea48da980013632cae016af476f8981a8f14683\"),",
" (1000000, 499985, 53478, \"fb3db7328c0c940f68d88e873c6554c9ec65616c825f5b933fd2e55e492e1be2\",",
" \"37e6b0aec8c146fa82e6e8d0eb776dbb1504fb2fff80e5fa74bff8eaddfee951\"),",
"]",
"",
"def primes_through(limit):",
" sieve = bytearray(b\"\\x01\") * (limit + 1)",
" sieve[:2] = b\"\\x00\\x00\"",
" for p in range(2, isqrt(limit) + 1):",
" if sieve[p]:",
" sieve[p*p:limit+1:p] = b\"\\x00\" * ((limit-p*p)//p+1)",
" return [p for p in range(2, limit + 1) if sieve[p]]",
"",
"def factorial_valuation(n, p):",
" value = 0",
" while n:",
" n //= p",
" value += n",
" return value",
"",
"def certify(n, k, expected_terms, expected_factor_sha, expected_tau_sha):",
" factors = []",
" tau = 1",
" histogram = {}",
" for p in primes_through(n):",
" exponent = factorial_valuation(n,p)-factorial_valuation(k,p)-factorial_valuation(n-k,p)",
" assert exponent >= 0",
" if exponent:",
" factors.append((p,exponent))",
" tau *= exponent+1",
" histogram[exponent] = histogram.get(exponent,0)+1",
" factor_data = \"*\".join(f\"{p}^{e}\" for p,e in factors)",
" tau_text = str(tau)",
" assert len(factors) == expected_terms",
" assert sha256(factor_data.encode()).hexdigest() == expected_factor_sha",
" assert sha256(tau_text.encode()).hexdigest() == expected_tau_sha",
" return {",
" \"n\":n, \"k\":k, \"factor_terms\":len(factors),",
" \"exponent_histogram\":sorted(histogram.items()),",
" \"factor_sha256\":expected_factor_sha, \"tau_digits\":len(tau_text),",
" \"tau_prefix\":tau_text[:32], \"tau_suffix\":tau_text[-32:],",
" \"tau_sha256\":expected_tau_sha,",
" }",
"",
"report = [certify(*case) for case in CASES]",
"payload = dumps(report,sort_keys=True,separators=(\",\",\":\"))",
"assert sha256(payload.encode()).hexdigest() == \"fea7d94ea1b9b59e253076f13b8caf82c47db0d875c7e0b71d351554ee7eb723\"",
"print(payload)"
],
"missing": [
"command",
"expected_output"
]
},
"formal_statement": null,
"source": {
"url": "https://doi.org/10.1134/S0001434613010331",
"locator": "Inline CPython standard-library computation reproduced on 2026-07-25"
},
"models": [],
"relations": [
{
"slug": "R68",
"title": "The exact record through n=20,000 occurs at (19,971, 9,949)",
"object_type": "claim",
"relation": "evidences",
"direction": "outgoing"
},
{
"slug": "R67",
"title": "A certified cutoff-scale coefficient gives a 16,113-digit lower bound",
"object_type": "claim",
"relation": "evidences",
"direction": "outgoing"
},
{
"slug": "binomial-divisor-record-1e6",
"title": "binomial divisor record 1e6",
"object_type": "problem",
"relation": "recorded_for",
"direction": "outgoing"
}
]
}8Provenance
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