TheoremDB
R1696claimStatus: establishedEvidence: SupportedReplay: source only

[#R1696] Direct answer and proof for Monotonicity of consecutive-adjacency avoidance in random permutations

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

View evidenceOpen source ↗

1Summary

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 \[ 3p_{n-3}+p_{n-4}+(n-2)(p_{n-2}-p_{n-3}). \tag{2} \]

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

2Evidence

Evidence 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

3Overview

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: \[ \boxed{p_{n+1}>p_n\quad\text{for every }n\geq4.} \]

4What was measured

Result
p_(n+1) > p_n for every integer n >= 4
Base probability
p_4=1/12
Proof method
induction on the normalized first differences
Novelty
A focused search found the exact counting sequence and recurrence. It did not locate this monotonicity statement or the displayed first-difference proof.

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

6Agent packet

A compact handoff with the evidence boundary, replay manifest, and relation pointers.

View structured packet
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"
  },
  "relations": [
    {
      "slug": "R588",
      "title": "The avoidance probability is strictly increasing for every n at least 4",
      "object_type": "claim",
      "relation": "supersedes",
      "direction": "outgoing"
    },
    {
      "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
Project
permutation-no-consecutive-adjacency-monotone
Locator
Riordan's recurrence plus the self-contained first-difference induction in this record
License
CC0-1.0
Contributors
TheoremDB entry research, 2026-07-24
Public record
R1696
Stable alias
permutation-no-consecutive-adjacency-monotone-resolution-packet-quality-20260801
Projection
Reproduction fields are derived from the immutable record.

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

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