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Actual Cloitre Weighted Cone

Abstract

A weighted cone bounds every actual upper-anchor deficit at least five under the complete conditional source hypotheses.

F is the Fibonacci sequence with F(0)=0 and F(1)=1. C is the actual positive-index Cloitre sequence with C(1)=C(2)=1. Its legal domain is D(N)=[1,N-1], its inner map is T(N,x)=N-C(x), and its orbit X(N,i) starts at N-1. The prescribed depth is d(N)=C(N-1), the selected point is g(N)=X(N,d(N)), and C(N)=C(g(N))+C(N-g(N)) for N>=3. Put Q(m,t)=F(m-1)-C(F(m)-t) on the full natural closed block 0<=t<=F(m-2). The upper cap makes Q the exact nonnegative integer difference. It is distinct from the golden excess C(n)-G(n).

Theorem 1.1 (Weighted cone on every natural closed block).

Lean statement: D5/S1/Recurrence/Invariants/CloitreActualWeightedCone.full30_6

Proof. Machine-checked in Lean as D5/S1/Recurrence/Invariants/CloitreActualWeightedCone.full30_6 (✓ std3). ∎

Source. Repository-derived.

Commentary.

For every U satisfying Hyp24_1(U), retain its complete inherited Hyp21_1: the finite ratio condition 22877C(n)<=15225n for 16384<=n<=131071; the full golden base and equality classification for 1<=n<=65535; and prescribed periodic entry for 3<=N<=52. The golden base states G(n)<=C(n), with equality implying n=F(j) or F(j)+1 for some j>=2, n+1=F(j) for an odd j>=3, or n in {11,24,25,59}. The global conditions include 1<=C(n), G(n)<=C(n)<=U(n)<=n, U(1)=1, the piecewise formula U(n)=min(n-F(j-2),F(j)) on F(j)<=n<F(j+1) for j>=3, and monotonicity and increments U(n)<=U(n+1)<=U(n)+1 on positive indices. For j>=2, U(F(j))=C(F(j))=G(F(j))=F(j-1); for j>=3, C(F(j)+1)=G(F(j)+1)=F(j-1)+1. For every q>=6 and t>=0, the right collar [F(q-1),F(q-1)+t] is legal and invariant under T(F(q)+t), captures every legal orbit, and contains every legal periodic point. For each N>=3, the earliest periodic entry of X(N,i) precedes or equals d(N). Hyp24_1 also includes C(F(j)-1)=F(j-1) for j>=5, negative-collar invariance and capture for j>=6 and b<=F(j-1), and the full legal periodic intersection between max(F(j-1),F(j)-b) and min(F(j),F(j)+F(j-3)-b).

Two additional finite full-block conditions are required. For every v<=F(18), Q(20,v)<=2 exactly when v<=35; when v>35, 3<=Q(20,v)<=max(3,v-36). For every v<=F(19), Q(21,v)<=3 exactly when v<=45; when v>45, 4<=Q(21,v)<=max(4,v-46). These conditions and the inherited foundations are premises; no instance of them is asserted.

For all natural m>=21 and all 0<=t<=F(m-2), Q(m,t)>=5 implies t-(3/2)Q(m,t)>=4m-81 over the rationals. Multiplication by two gives the equivalent subtraction-free integer inequality 3Q(m,t)+8m<=2*t+162.

Let Z(m)=3*m+floor((m-1)/3)-24. Periodic predecessors and the finite order-twenty shelf give Q(21,v)=3 on v=42..45. An induction propagates Q(m,v)<=3 exactly on v<=Z(m), the exterior shelf, and the last four values equal to three. At the two-point boundary the adjacent actual row fixes the depth; exterior alternation from N-1 fixes its parity-selected phase.

The golden estimate 8Q(m,q)<=5q+8 and the zero and exterior shelves prove q-(3/2)*Q(m,q)>=4 for every legal q>=4, using the ranges 4..11, 12..50, 51..87 and q>=88. Both orders twenty-one and twenty-two start the weighted induction. At higher orders the same actual children have inherited orders m-1,m-2 and legal gaps z,w with z+w=t. The cases are first deficit at least five, first deficit four with second deficit one through four, and first deficit four with second deficit at least five. No monotonicity of C, independent child choices, free periodic phase or limiting ratio is required.

References