TheoremDB

Problem packetResearch packetR1500

R1500Reproduced evidence

Direct answer and proof for Longest four-number Ducci trajectory in the 100 box

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Authored summary

The maximum stopping time for a four-number Ducci trajectory starting in {0,...,100}^4 is 13, and exactly 2,816 ordered quadruples attain it.

The recorded result has been reproduced within its stated scope.

Recorded status: established

Recorded scope: all ordered quadruples in {0,...,100}^4 under the four-number Ducci map

Complete recorded scope and conditions
{
  "kind": "bounded",
  "statement": "all ordered quadruples in {0,...,100}^4 under the four-number Ducci map",
  "bounds": {
    "coordinate": {
      "min": 0,
      "max": 100
    },
    "ordered_quadruples": {
      "min": 104060401,
      "max": 104060401
    }
  },
  "exhaustive": true
}

Originating problem: Longest four-number Ducci trajectory in the 100 box

Authored record and scope
Authored title
Direct answer and proof for Longest four-number Ducci trajectory in the 100 box
Record type
claim
Stored status
established
Evidence grade
reproduced
Recorded scope data
{ "kind": "bounded", "statement": "all ordered quadruples in {0,...,100}^4 under the four-number Ducci map", "bounds": { "coordinate": { "min": 0, "max": 100 }, "ordered_quadruples": { "min": 104060401, "max": 104060401 } }, "exhaustive": true }

2Authored explanation

Write \(\tau(x)\) for the first time at which \(D^\tau(x)=0\). Exact enumeration gives \[ \max_{x\in\{0,\ldots,100\}^4}\tau(x)=13. \] One witness is \((0,7,20,44)\). Its trajectory is \[ \begin{aligned} &(0,7,20,44),(7,13,24,44),(6,11,20,37),(5,9,17,31),\\ &(4,8,14,26),(4,6,12,22),(2,6,10,18),(4,4,8,16),\\ &(0,4,8,12),(4,4,4,12),(0,0,8,8),(0,8,0,8),\\ &(8,8,8,8),(0,0,0,0). \end{aligned} \] The candidate witness \((57,81,37,44)\) belongs to the same class: subtracting 37 gives \((20,44,0,7)\), a rotation of the displayed witness.

The number of ordered starting tuples at each stopping time \(0,1,\ldots,13\) is \[ (1,100,20100,1353400,53030200,17977176,22886192,5951312,2093072,540976,165040,31136,8880,2816). \] These counts sum to \(101^4=104060401\).

The displayed trajectory from (0,7,20,44) takes thirteen Ducci steps, which proves the lower bound. For the upper bound, normalize each nonconstant quadruple by subtracting its smallest coordinate, dividing by the positive greatest common divisor of its coordinates, and taking the least rotation or reflection. Translation, positive scaling, rotation, and reflection preserve the stopping time, so this assigns every nonconstant quadruple to one primitive canonical representative. If that representative has maximum coordinate m and dihedral orbit size o, the number of tuples in the box that it represents is o times the sum of 101-gm over positive integers g with gm at most 100. The companion C17 certificate enumerates all 425,479 primitive representatives, computes this weight for each one, and handles the 101 constant tuples separately. The weights sum to 101^4, so the classes exhaust the box without omission. The last nonzero stopping-time class is thirteen. Together with the displayed witness, this proves that the exact maximum is thirteen.

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Replay material: source only

3Evidence

Replay package: source only

A verification source is cited. This record has no executable replay attached.

Verification source: oeis.org ↗, Independent exhaustive artifact dfb100-artifact-symmetry-quotient-enumeration; OEIS A065677 gives the same bounded maximum sequence

4What was measured

Proof remediation

reviewed on2026-08-01predecessordfb100-claim-exact-maximumchangeDirect-answer summary and readable proof presentation.

5How it connects

Replaces

Recorded for

Machine-readable record

Copy the structured record when continuing this work with an agent.

json
{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R1500",
  "content_hash": null,
  "slug": "ducci-four-box-100-resolution-packet-quality-20260801",
  "type": "claim",
  "title": "Direct answer and proof for Longest four-number Ducci trajectory in the 100 box",
  "summary": "The maximum stopping time for a four-number Ducci trajectory starting in {0,...,100}^4 is 13, and exactly 2,816 ordered quadruples attain it.",
  "relevance": "For Longest four-number Ducci trajectory in the 100 box, this successor supplies the direct answer and complete human-readable proof required by the Resolution presentation.",
  "relevance_source": "recorded",
  "body": "Write \\(\\tau(x)\\) for the first time at which \\(D^\\tau(x)=0\\). Exact enumeration gives\n\\[\n\\max_{x\\in\\{0,\\ldots,100\\}^4}\\tau(x)=13.\n\\]\nOne witness is \\((0,7,20,44)\\). Its trajectory is\n\\[\n\\begin{aligned}\n&(0,7,20,44),(7,13,24,44),(6,11,20,37),(5,9,17,31),\\\\\n&(4,8,14,26),(4,6,12,22),(2,6,10,18),(4,4,8,16),\\\\\n&(0,4,8,12),(4,4,4,12),(0,0,8,8),(0,8,0,8),\\\\\n&(8,8,8,8),(0,0,0,0).\n\\end{aligned}\n\\]\nThe candidate witness \\((57,81,37,44)\\) belongs to the same class: subtracting 37 gives \\((20,44,0,7)\\), a rotation of the displayed witness.\n\nThe number of ordered starting tuples at each stopping time \\(0,1,\\ldots,13\\) is\n\\[\n(1,100,20100,1353400,53030200,17977176,22886192,5951312,2093072,540976,165040,31136,8880,2816).\n\\]\nThese counts sum to \\(101^4=104060401\\).\n\nThe displayed trajectory from (0,7,20,44) takes thirteen Ducci steps, which proves the lower bound. For the upper bound, normalize each nonconstant quadruple by subtracting its smallest coordinate, dividing by the positive greatest common divisor of its coordinates, and taking the least rotation or reflection. Translation, positive scaling, rotation, and reflection preserve the stopping time, so this assigns every nonconstant quadruple to one primitive canonical representative. If that representative has maximum coordinate m and dihedral orbit size o, the number of tuples in the box that it represents is o times the sum of 101-gm over positive integers g with gm at most 100. The companion C17 certificate enumerates all 425,479 primitive representatives, computes this weight for each one, and handles the 101 constant tuples separately. The weights sum to 101^4, so the classes exhaust the box without omission. The last nonzero stopping-time class is thirteen. Together with the displayed witness, this proves that the exact maximum is thirteen.",
  "status": "established",
  "evidence_grade": "reproduced",
  "scope": {
    "kind": "bounded",
    "statement": "all ordered quadruples in {0,...,100}^4 under the four-number Ducci map",
    "bounds": {
      "coordinate": {
        "min": 0,
        "max": 100
      },
      "ordered_quadruples": {
        "min": 104060401,
        "max": 104060401
      }
    },
    "exhaustive": true
  },
  "reproduction": {
    "schema": "theoremdb-reproduction-v1",
    "readiness": "source_only",
    "kind": "claim",
    "citation": {
      "url": "https://oeis.org/A065677",
      "locator": "Independent exhaustive artifact dfb100-artifact-symmetry-quotient-enumeration; OEIS A065677 gives the same bounded maximum sequence"
    },
    "missing": [
      "source",
      "command",
      "runtime",
      "expected_output"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": "https://oeis.org/A065677",
    "locator": "Independent exhaustive artifact dfb100-artifact-symmetry-quotient-enumeration; OEIS A065677 gives the same bounded maximum sequence"
  },
  "models": [],
  "continuation": null,
  "relations": [
    {
      "slug": "R211",
      "title": "The longest trajectory in the 100 box has 13 steps",
      "object_type": "claim",
      "relation": "supersedes",
      "direction": "outgoing",
      "metadata": {
        "reason": "Adds the direct answer and human-readable proof reviewed on 2026-08-01."
      }
    },
    {
      "slug": "ducci-four-box-100",
      "title": "ducci four box 100",
      "object_type": "problem",
      "relation": "recorded_for",
      "direction": "outgoing"
    }
  ]
}

7Provenance

View source, identifiers, and projection details

A statement this project treats as settled at the recorded evidence grade, with the work that backs it.

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