The Eta Bridge

Phi_3(2) = 2^2 + 2 + 1 = 7

The modular forms page showed WHAT: tau(p) mod 23 classifies primes into inner and outer classes. This page answers WHY. The Cyclotomic Period Theorem traces periods 2 and 3 back to the Cunningham chain. Then the Smooth Zone Ladder shows the partition function's smooth zone.

The Cyclotomic Period Theorem

For any prime p (except 23 = 2*11+1), tau(p^k) mod 23 cycles with period depending on whether p is a quadratic residue (inner, period 3) or non-residue (outer, period 2) mod 23. These periods are forced:

Cyclotomic Period Derivation

Step 1: 11 = 1+2+3+5. Step 2: 23 = 2*11+1 (Cunningham step). So 23 mod 3 = 2 (not 1). Step 3: Since 23 mod 3 != 1, the 3rd cyclotomic polynomial Phi_3(x) = x^2+x+1 is irreducible mod 23. Step 4: For QR primes: characteristic polynomial = Phi_3 => roots in F_{23^2} with order 3 => period 3. Step 5: For NR primes: char poly = x^2-1 = (x-1)(x+1) splits => period 2.

3 appears three ways: as the period, as the cyclotomic index, and as the class number h(-23). A self-referential loop.

Cyclotomic Polynomials at D = 2

Evaluating cyclotomic polynomials at x = 2 generates the ring's primes. Not chosen -- forced by algebra:

Polynomial	Formula at x=2	Value	Note
Phi_1(2)	2 - 1	1	Identity
Phi_2(2)	2 + 1	3	Prime
Phi_3(2)	4 + 2 + 1	7	Prime
Phi_6(2)	4 - 2 + 1	3	Same as Phi_2
Product	Phi_1Phi_2Phi_3*Phi_6	63	9 * 7

Phi_3(2) = 2^2+2+1 = 7. The third cyclotomic at x=2 yields 7. Phi_2(2) = 3. Phi_4(2) = 5. Phi_10(2) = 11. Phi_12(2) = 13. The ring's primes appear at cyclotomic orders that divide 12.

The Smooth Zone Ladder

For the initial prime segment S_k = {p_1, ..., p_k}, define B(S_k) = the largest n such that ALL partition values p(1),...,p(n) are S_k-smooth. Watch how the smooth zone grows:

Smooth set S	B(S)	Breaker	Jump
{2}	2	p(3) = 3	+1
{2, 3}	3	p(4) = 5	+1
{2, 3, 5}	4	p(5) = 7	+1
{2, 3, 5, 7}	5	p(6) = 11	+1
{2,3,5,7,11}	12	p(13) = 101	+7 jump!

Smooth Zone Jump Theorem

For k = 1..4, B(S_k) = k+1 (grows by 1 each time). Adding 11 causes a jump of 7, from B=5 to B=12. Why? Because p(6) = 11, so adding 11 immediately covers the breaker, and p(7) through p(12) are all composites with factors in {2,3,5,7,11}. The breaker at position 13 is p(13) = 101, which has a prime factor outside the set.

Second smooth block: p(14) through p(19) are all {2,3,5,7,11}-smooth again. Length = 6. Breaker = p(20) contains factor 19. Total smooth = 12 + 6 = 18.

Partition Values ARE Axiom Primes

The partition values p(2) through p(6) are exactly the primes {2, 3, 5, 7, 11}: p(2)=2, p(3)=3, p(4)=5, p(5)=7, p(6)=11. Each smooth set S_k breaks at p(k+2) because the partition function equals the prime sequence at these positions. The breaker IS the next prime.

The Thread

Three threads converge on the same numbers:

Eta function

eta^24 = Delta, weight 12

Unique cusp form. 24 = 8 * 3.

Partition function

{2,3,5,7,11}-smooth for n <= 12

Breaks at position 13. Ramanujan congruences at exactly {5, 7, 11}.

Class numbers

h(-23) = 3

The class number IS the cyclotomic period.

Why 24?

8 * 3 = 24

Leech lattice (24 dimensions). eta^24 = Delta.

Eta, partitions, class numbers, cyclotomic polynomials -- four branches of number theory, all governed by the same small primes.

Contrast Table

Cyclotomic periodsTools of algebraic number theoryPeriods 2 and 3 govern tau(p^k) mod 23. Forced by 23 mod 3 = 2.Partition smooth zonePartition function studied without reference to smooth sets{2,3,5,7,11}-smooth for n <= 12. Jump of 7 when adding 11. Breaks at position 13.eta^24 = DeltaModular form with weight 12Weight 12 = smooth zone boundary = Carmichael lambda of Z/210.Ramanujan congruencesCongruences at 5, 7, 11. Why these primes?Exactly {5, 7, 11} = the outer class (non-residues mod 23).h(-23) = 3Class number computation23 = 2*11+1. The class number of Q(sqrt(-23)) equals the cyclotomic period.

Explore: Cyclotomic at D=2

Enter n (1-12) to compute the n-th cyclotomic polynomial at x=2. Watch the primes appear: Phi_1=1, Phi_2=3, Phi_3=7, Phi_4=5, Phi_10=11, Phi_12=13.

Order n:

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