Catalan convolution and a divisibility certificate

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

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derivation-mko-catalan-numbers-01 · ocme-new142-reader-adoption-20261021

Goal and scope

Generate C4 from C0 and check both closed forms.

An exact n=4 instance and general integrality identity, not every Catalan bijection.

Givens and symbols

  • work
  • G1 — C0=1 and for every n≥0 use C_(n+1)=sum_(i=0)^n C_i C_(n-i). Compare the two cited Catalan closed forms at n=4.
  • C_n — zero-based Catalan sequence

Rules and sources

  • mko_definition · cited — C_(n+1)=sum over i=0..n of C_i*C_(n−i), and C_n=binomial(2n,n)/(n+1)=binomial(2n,n)−binomial(2n,n+1). Before division, certify that n+1 divides the central binomial coefficient; integer truncation is not a substitute. The domain is n≥0 and the C_0/C_1 indexing conventions must not be mixed. · mko-catalan-numbers
  • work · prerequisite — Finite convolution definition; binomial factorial identities and comb=0 beyond its upper index; the Catalan closed forms are cited MKO facts, not general theorems proved by this example.

Main derivation

S01 · implies

Before

C_0=1

After

C_1=C_0C_0=1

Conditions: none

Rule: work · Uses: G1, work

Leaf basis: work

Fix zero-based indexing first rather than mixing C1 conventions.

S02 · implies

Before

C_2=C_0C_1+C_1C_0

After

C_2=2

Conditions: none

Rule: work · Uses: G1, S01, work

Leaf basis: work

At n=1 include both splits.

S03 · implies

Before

C_3=C_0C_2+C_1C_1+C_2C_0

After

C_3=2+1+2=5

Conditions: none

Rule: work · Uses: G1, S02, work

Leaf basis: work

List all three splits at n=2.

S04 · implies

Before

C_4=C_0C_3+C_1C_2+C_2C_1+C_3C_0

After

C_4=5+2+2+5=14

Conditions: none

Rule: work · Uses: G1, S03, work

Leaf basis: work

All four splits at n=3 give 14.

S05 · implies

Before

\binom84=8!/(4!4!)

After

\binom84=70,\quad70\bmod5=0

Conditions: none

Rule: work · Uses: G1, S04, work

Leaf basis: work

Compute the central binomial coefficient and certify remainder zero before integer division.

S06 · implies

Before

\binom84/(4+1)

After

70/5=14=C_4

Conditions: none

Rule: work · Uses: G1, S05, work

Leaf basis: work

After divisibility, check the first closed form at n=4.

S07 · implies

Before

\binom85=56

After

\binom84-\binom85=70-56=14

Conditions: none

Rule: work · Uses: G1, S06, work

Leaf basis: work

The second closed form gives the same integer as a cross-check.

S08 · implies

Before

\binom{2n}{n+1}=\binom{2n}{n}\frac{n}{n+1}\quad(n\ge0)

After

\binom{2n}{n}-\binom{2n}{n+1}=\frac1{n+1}\binom{2n}{n}\in\mathbb Z

Conditions: none

Rule: work · Uses: G1, S07, work

Leaf basis: work

The factorial ratio gives the identity; an integer difference proves divisibility, with comb(0,1)=0 at n=0.

Substeps and lemmas

No substeps in this guide.

Reverse audits

  • S08 · premise_audit · G1, S07, work — This proves integrality, not the entire general closed-form solution of the convolution.
  • S07 · premise_audit · G1, S06, work — The lower index n+1=5 must not be replaced by n-1.
  • S06 · premise_audit · G1, S05, work — This numerical agreement alone does not prove the closed form for every n.
  • S05 · premise_audit · G1, S04, work — A truncated quotient is not a divisibility certificate.
  • S04 · premise_audit · G1, S03, work — Adding only endpoints would give the wrong value 10.
  • S03 · premise_audit · G1, S02, work — Do not omit the middle split.
  • S02 · premise_audit · G1, S01, work — Equal terms still have separate indices.
  • S01 · premise_audit · G1, work — C0 for the empty object is not zero.

Error controls and negative controls

Index

C4 uses n=3 in the convolution and n=4 in the closed form.

Exact division

Check70 mod5=0 before 70/5=14.

  • C0=0 gives the same sequence. → reject — The first convolution would give C1=0 rather than 1.
  • 70//5 replaces the general remainder obligation. → reject — Truncation itself proves no divisibility; remainder zero must be checked.
  • C4 only needs C0C3+C3C0=10. → reject — The two middle splits each add2, giving 14.

Result and limitations

C4=14, central coefficient 70 is divisible by 5, and 70-56=14.

    Verification and source binding

    4 declared checks for this guide; only the listed expressions are covered.

    • source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
    • adopted source SHA-256: 7522ef04029fd7f26aab4081199d43b82a2d2f70998d7d556d7c788ecb395599
    • original source SHA-256: ed47d97d69cfdf0cb831c56ec2b237fbcd2f49986cf1490c20647d56623f8552
    • Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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