TheoremDB

Problem packetResearch packetR1696

R1696Sourced evidence

Direct answer and proof for Monotonicity of consecutive-adjacency avoidance in random permutations

View evidenceOpen source ↗
Link to a section

Authored summary

For every integer n at least 4, p_(n+1) is strictly greater than p_n, so the avoidance probability is strictly increasing.

The record cites sources for its explanation.

Recorded status: established

Recorded scope: every comparison p_(n+1) > p_n for n at least 4

Complete recorded scope and conditions
{
  "kind": "universal",
  "statement": "every comparison p_(n+1) > p_n for n at least 4"
}

Originating problem: Monotonicity of consecutive-adjacency avoidance in random permutations

Authored record and scope
Authored title
Direct answer and proof for Monotonicity of consecutive-adjacency avoidance in random permutations
Record type
claim
Stored status
established
Evidence grade
sourced
Recorded scope data
{ "kind": "universal", "statement": "every comparison p_(n+1) > p_n for n at least 4" }

2Authored explanation

Let \(a_n\) count the admissible permutations and put \(p_n=a_n/n!\). The exact recurrence in the companion claim is \[ a_n=(n+1)a_{n-1}-(n-2)a_{n-2}-(n-5)a_{n-3}+(n-3)a_{n-4}\qquad(n\geq4), \] with \(a_0=a_1=1\) and \(a_2=a_3=0\).

Set \(\Delta_n=p_n-p_{n-1}\). Dividing the recurrence by \(n!\), subtracting \(p_{n-1}\), and collecting terms gives \[ n\Delta_n=\Delta_{n-1}+ \frac{(n-2)p_{n-2}-(n-5)p_{n-3}+p_{n-4}} {(n-1)(n-2)}. \tag{1} \] The numerator in the fraction has the useful form \[ 3p_{n-3}+p_{n-4}+(n-2)(p_{n-2}-p_{n-3}). \tag{2} \]

Now \(p_2=p_3=0\) and \(p_4=2/4!=1/12\), so \(\Delta_4>0\). For \(n=5\), the comparison needed in (2) is \(p_3\geq p_2\). At every later step it follows from the preceding induction cases. All probabilities are nonnegative, so (2) is nonnegative. Equation (1) then gives \(\Delta_n>0\) from \(\Delta_{n-1}>0\). Induction proves \(p_n>p_{n-1}\) for every \(n\geq4\). Replacing \(n\) by \(n+1\) answers the stated question: \[ \boxed{p_{n+1}>p_n\quad\text{for every }n\geq4.} \]

Continue this work
Replay material: source only

3Evidence

Replay package: source only

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

Verification source: doi.org ↗, Riordan's recurrence plus the self-contained first-difference induction in this record

4What was measured

Recurrence source

authorJohn RiordantitleA recurrence for permutations without rising or falling successionsvenueAnnals of Mathematical Statistics 36 (1965), 708-710doi10.1214/aoms/1177700181

Proof remediation

reviewed on2026-08-01predecessorpnc-claim-strict-probability-monotonicitychangeDirect-answer summary and readable proof presentation.

5How it connects

Recorded for

Machine-readable record

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

json
{
  "schema": "theoremdb-agent-record-v1",
  "ref": "R1696",
  "content_hash": null,
  "slug": "permutation-no-consecutive-adjacency-monotone-resolution-packet-quality-20260801",
  "type": "claim",
  "title": "Direct answer and proof for Monotonicity of consecutive-adjacency avoidance in random permutations",
  "summary": "For every integer n at least 4, p_(n+1) is strictly greater than p_n, so the avoidance probability is strictly increasing.",
  "relevance": "For Monotonicity of consecutive-adjacency avoidance in random permutations, this successor supplies the direct answer and complete human-readable proof required by the Resolution presentation.",
  "relevance_source": "recorded",
  "body": "Let \\(a_n\\) count the admissible permutations and put \\(p_n=a_n/n!\\). The exact recurrence in the companion claim is\n\\[\na_n=(n+1)a_{n-1}-(n-2)a_{n-2}-(n-5)a_{n-3}+(n-3)a_{n-4}\\qquad(n\\geq4),\n\\]\nwith \\(a_0=a_1=1\\) and \\(a_2=a_3=0\\).\n\nSet \\(\\Delta_n=p_n-p_{n-1}\\). Dividing the recurrence by \\(n!\\), subtracting \\(p_{n-1}\\), and collecting terms gives\n\\[\nn\\Delta_n=\\Delta_{n-1}+\n\\frac{(n-2)p_{n-2}-(n-5)p_{n-3}+p_{n-4}}\n{(n-1)(n-2)}. \\tag{1}\n\\]\nThe numerator in the fraction has the useful form\n\\[\n3p_{n-3}+p_{n-4}+(n-2)(p_{n-2}-p_{n-3}). \\tag{2}\n\\]\n\nNow \\(p_2=p_3=0\\) and \\(p_4=2/4!=1/12\\), so \\(\\Delta_4>0\\). For \\(n=5\\), the comparison needed in (2) is \\(p_3\\geq p_2\\). At every later step it follows from the preceding induction cases. All probabilities are nonnegative, so (2) is nonnegative. Equation (1) then gives \\(\\Delta_n>0\\) from \\(\\Delta_{n-1}>0\\). Induction proves \\(p_n>p_{n-1}\\) for every \\(n\\geq4\\). Replacing \\(n\\) by \\(n+1\\) answers the stated question:\n\\[\n\\boxed{p_{n+1}>p_n\\quad\\text{for every }n\\geq4.}\n\\]",
  "status": "established",
  "evidence_grade": "sourced",
  "scope": {
    "kind": "universal",
    "statement": "every comparison p_(n+1) > p_n for n at least 4"
  },
  "reproduction": {
    "schema": "theoremdb-reproduction-v1",
    "readiness": "source_only",
    "kind": "claim",
    "citation": {
      "url": "https://doi.org/10.1214/aoms/1177700181",
      "locator": "Riordan's recurrence plus the self-contained first-difference induction in this record"
    },
    "missing": [
      "source",
      "command",
      "runtime",
      "expected_output"
    ]
  },
  "formal_statement": null,
  "source": {
    "url": "https://doi.org/10.1214/aoms/1177700181",
    "locator": "Riordan's recurrence plus the self-contained first-difference induction in this record"
  },
  "models": [],
  "relations": [
    {
      "slug": "R588",
      "title": "The avoidance probability is strictly increasing for every n at least 4",
      "object_type": "claim",
      "relation": "supersedes",
      "direction": "outgoing",
      "metadata": {
        "reason": "Adds the direct answer and human-readable proof reviewed on 2026-08-01."
      }
    },
    {
      "slug": "permutation-no-consecutive-adjacency-monotone",
      "title": "permutation no consecutive adjacency monotone",
      "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.

Sign in to follow

Sign in in another tab, then return here.

Open sign-in in another tab

Report a problem

Report location:

Your ChatGPT account

Opening ChatGPT

ChatGPT is opening in a new tab.