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.
derivation-mko-catalan-numbers-01 · ocme-new142-reader-adoption-20261021
Generate C4 from C0 and check both closed forms.
An exact n=4 instance and general integrality identity, not every Catalan bijection.
workG1 — 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 sequencemko_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-numberswork · 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.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.
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.
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.
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.
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.
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.
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.
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 ZConditions: 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.
No substeps in this guide.
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.C4 uses n=3 in the convolution and n=4 in the closed form.
Check70 mod5=0 before 70/5=14.
reject — The first convolution would give C1=0 rather than 1.reject — Truncation itself proves no divisibility; remainder zero must be checked.reject — The two middle splits each add2, giving 14.C4=14, central coefficient 70 is divisible by 5, and 70-56=14.
4 declared checks for this guide; only the listed expressions are covered.
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