The Smooth Census

run(k) = 13 - k

An integer is 11-smooth if its only prime factors come from {2, 3, 5, 7, 11}. For n <= 12, every binomial coefficient C(n,k) is 11-smooth. At n = 13, the prime 13 enters every C(13,k) for 0 < k < 13. Seven independent classical sequences confirm the same boundary: 13 is the universal wall.

The Binomial Smooth Theorem

For n <= 12, the factorial n! contains only primes <= 11. So C(n,k) = n!/(k!(n-k)!) is a ratio of 11-smooth numbers, hence 11-smooth. At n = 13, the prime 13 enters 13! and divides C(13,k) for all 0 < k < 13 (by Lucas theorem: 13 is prime, so 13 | C(13,k)).

Binomial Smooth Theorem (PROVED)

C(n,k) is 11-smooth for all k <= n <= 12. At n = 13, every C(13,k) = 13 * (smooth number). The smooth run length at depth k is run(k) = 13 - k.

k	Run = 13-k	C(13,k)/13
1	12	1
2	11	6 = 2*3
3	10	22 = 2*11
4	9	55 = 5*11
5	8	99 = 9*11
6	7	132 = 4311
7	6	132 = 4311
8	5	99 = 9*11
9	4	55 = 5*11
10	3	22 = 2*11
11	2	6 = 2*3
12	1	1

Run lengths are {12, 11, 10, ..., 1}. The quotients C(13,k)/13 form a palindrome: {1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1}. Every quotient is 11-smooth. Pascal's triangle is smooth below row 13.

The Independent Smooth Census

Classical sequences whose smooth runs are independent of binomial coefficients. Each has an initial 11-smooth run, then 13 appears:

Sequence	Smooth run	First non-smooth
Bernoulli denom B_2n	5	denom(B_12) = 2730 = 235713
Fibonacci F(n)	6	F(7) = 13
Catalan C_n	6	C_7 = 429 = 31113
Sum of divisors sigma(n)	8	sigma(9) = 1+3+9 = 13
Triangle T(n)	11	T(12) = 78 = 2313
Partition p(n)	12	p(13) = 101 (prime > 11)
Bell B(n)	5	B(5) = 52 = 4*13

In 6 of 7 sequences, 13 appears in the blocking value itself. In partitions, 13 is the blocking position: p(13) = 101 is the first non-smooth partition number.

How 13 Enters Each Sequence

Each entry mechanism is independent. Divisor sums, recurrences, combinatorial products, von Staudt-Clausen -- all different mathematics, same wall.

sigma(9) = 13

Sum of divisors

1 + 3 + 9 = 13. The divisors of 9 = 3^2 sum to exactly 13.

F(7) = 13

Fibonacci

The 7th Fibonacci number is 13. Cumulative addition: 1, 1, 2, 3, 5, 8, 13.

B(5) = 52

Bell numbers

52 = 4 * 13. The 5th Bell number contains 13 as a factor.

p(13) = 101

Partitions

The 13th partition number is 101, a prime larger than 11.

The Smooth Ladder

Ordering by run length, the sequences form a ladder from 5 to 12. The distinct run lengths are {5, 6, 8, 11, 12}. Gaps between rungs: 1, 2, 3, 1.

Run = 5

Bernoulli, Bell

Two sequences share this rung.

Run = 6

Fibonacci, Catalan

F(7) = 13 directly; C_7 = 429 = 3*11*13.

Run = 8

sigma(n)

The divisor sum function.

Run = 11

Triangle numbers

T(n) = C(n+1,2) is a binomial coefficient.

Run = 12

Partition function

13 is the blocking position, not value.

Combined

All integers 1..12 appear as run lengths

The binomial theorem contributes every value 1..12. The seven sequences sample 5 distinct rungs from that range.

Explore: Smooth Checker

Enter any positive integer. See its factorization into {2, 3, 5, 7, 11} and whether it is 11-smooth.

Check smoothness of n:

Try: 13 (first non-smooth prime), 42, 420, 2310 = 2*3*5*7*11, 970200, 12612600.

CRT Smooth Projection

Which CRT channels can contain non-axiom-smooth residues? (Axiom-smooth: all prime factors in {2, 3, 5, 7, 11, 13, 17}.) Only the fat channels Z/25 and Z/49. Every residue in Z/8, Z/9, Z/11, Z/13, and Z/17 is axiom-smooth -- all primes below 18 are axiom primes. Non-axiom primes (19, 23, 29, ...) can only appear in fat channels.

CRT Smooth Projection (PROVED)

Only fat channels Z/25 and Z/49 contain non-axiom-smooth residues. E-channel: 2 intruders {f(5) = 19, c(11) = 23}. b-channel: 10 non-smooth values (8 intruder primes + 2 composites). CRT-smooth fraction = 897/1225 = c(11)*3*13 / (25*49). Intruder product: 19*23 = 437 = lambda + ESCAPE. ADDRESS = 137 projects to chain elements: mod 30 = ESCAPE, mod 42 = L, mod 70 = SOUL. KEY = 41 is NOT CRT-smooth (b-channel intruder).

E-channel: 2 intruders

{f(5) = 19, c(11) = 23}

The depth quadratic and Cunningham map produce the only non-axiom-smooth residues in Z/25. 23 out of 25 residues are axiom-smooth.

b-channel: 10 intruders

8 primes + 2 composites in Z/49

8 primes > 17: {19, 23, 29, 31, 37, 41, 43, 47}. 2 composites: {38=2*19, 46=2*23}. 39 out of 49 residues are axiom-smooth.

897/1225 smooth

= 73.2%

CRT-smooth fraction: 23*39 / (25*49). The numerator 897 = c(11)*3*13 is itself axiom-smooth.

ADDRESS projects

137 mod 30 = 17, mod 42 = 11, mod 70 = 67

The prime 137 projects to ESCAPE, L, and SOUL through three chain-based moduli. Bookend: 137 mod 2 = 137 mod 17 = 1.

Contrast

Aspect	Standard view	Ring structure
Smooth runs	Scattered counts: 5, 6, 8, 11, 12	All five run lengths are chain primes or their products: 5, 6=23, 8=2^3, 11, 12=43
Blocker	13 is the next prime after 11	13 = 2^2 + 3^2, where {2,3,5,7,11} generate the ring Z/2,310
Entry pattern	Each sequence has its own mechanism	All mechanisms hit 13. Pascal row 13 assigns run lengths 12 down to 1
Ladder gaps	1, 2, 3, 1 -- small integers	Gaps 1, 2, 3, 1 -- only the three smallest positive integers appear