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Actual Cloitre Deficit-Four Positive Fees

Abstract

Positive fees on the full actual deficit-four support settle along the complete finite first-child spine.

F is the Fibonacci sequence with F(0)=0 and F(1)=1. Put phi=(1+sqrt(5))/2 and G(n)=floor((n+1)/phi). The actual sequence satisfies 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 X(N,i)=T(N)^i(N-1). The depth is d(N)=C(N-1), g(N)=X(N,d(N)), and C(N)=C(g(N))+C(N-g(N)) for N>=3. On the natural closed block 0<=t<=F(m-2), put Q(m,t)=F(m-1)-C(F(m)-t). The upper cap makes this an exact nonnegative integer difference. All roots and descendants use this same C, g, d and T.

Theorem 1.1 (Full-support positive-fee partition and complete finite spine settlement).

Lean statement: D5/S1/Recurrence/Invariants/CloitreActualDeficitFourFees.full30_5

Proof. Machine-checked in Lean as D5/S1/Recurrence/Invariants/CloitreActualDeficitFourFees.full30_5 (✓ std3). ∎

Source. Repository-derived.

Commentary.

Hyp24_1(U) includes the 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}. For every positive n, the global bounds are 1<=C(n) and G(n)<=C(n)<=U(n)<=n. The upper function satisfies 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, and for j>=6 and 0<=b<=F(j-1), with N=F(j+1)-b, the collar [F(j)-b,F(j)] intersected with D(N) is invariant under T(N) and captures every legal orbit. Every legal periodic point x of T(N) satisfies max(F(j-1),F(j)-b)<=x and x<=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.

Define Z(m)=3m+floor((m-1)/3)-24, B(m)=4m-34, and Phi(m,t)=max((t:Z)-(B(m):Z),0). Define K(N)=(g(N):Z)-(T(N,g(N)):Z), casting both terms before subtraction. Phi and K are signed integer expressions. For legal orders m>=21 the natural threshold B(m) agrees with the displayed subtraction. Define epsilon(m,b)=1 when b=Z(m)+1 and m mod 3!=1, and zero otherwise.

For every m>=22 and legal gap b<=F(m-2) with Q(m,b)=4, put N=F(m)-b, z=F(m-1)-g(N), and w=F(m-2)-(N-g(N)). The actual child coordinates satisfy g(N)=F(m-1)-z, N-g(N)=F(m-2)-w, z+w=b, z<=F(m-3), and w<=F(m-4). Their deficits are Q(m-1,z)=4 and Q(m-2,w)=0. The signed identities are K(N)=w-4 and b-B(m)=(z-B(m-1))+K(N). The positive fee is exactly max(K(N),0)=Phi(m,b)-Phi(m-1,z).

The low sector b<=B(m) satisfies Z(m)<b, z<=B(m-1), and K(N)<=0. Both potentials and the positive fee are zero. In its strict part b<B(m), the actual selected gaps are z=b-4+epsilon(m,b), w=4-epsilon(m,b), with K(N)=-epsilon(m,b) and z<B(m-1). Thus the exceptional epsilon=1 phase has complementary gap three. The separate endpoint b=B(m) has z=B(m-1), w=4 and K(N)=0.

The high sector b>B(m) satisfies z>=B(m-1), K(N)>=0 and w>=4. When K(N)=0, z=b-4>B(m-1). When K(N)>0, there is a positive period p of the actual selected g(N) under this same T(N). Take y=T(N)^(p-1)(g(N)) and u=F(m-1)-y. Then T(N,y)=g(N), y belongs to D(N), y=F(m-1)-u, u<=F(m-3), and u<=b-4. This physical predecessor satisfies z=b-Q(m-1,u), Q(m-1,u)=w>=5, and u-w>=4*m-42. Consequently z>=B(m-1). The predecessor comes from the same selected cycle, without an independently chosen phase or preimage.

For every M>=22 and legal deficit-four start b0, define N_s=g^s(F(M)-b0), m_s=M-s and b_s=F(m_s)-N_s. Then b_0=b0. Every node s<=M-21, including the terminal node, satisfies m_s>=21, N_s=F(m_s)-b_s, b_s<=F(m_s-2) and Q(m_s,b_s)=4. Every outgoing edge s=22 and N_(s+1)=g(N_s), m_(s+1)=m_s-1, b_(s+1)=F(m_s-1)-g(N_s). Its fee is Phi(m_s,b_s)-Phi(m_(s+1),b_(s+1)). The terminal order is exactly m_(M-21)=21.

Put S=sum over s in {0,…,M-22} of max(K(N_s),0), equivalently the integer sum over range(M-21). The exact settlement is S=Phi(M,b0)-Phi(21,b_(M-21)). The terminal enclosure is b_(M-21)<=487=348, since P(21)=floor((21-2)^2/3)-321+30=87. With B(21)=50, 0<=Phi(21,b_(M-21))<=298. Hence S>=0 and S<=Phi(M,b0)<=S+298. Also b0<=B(M)+Phi(M,b0)<=4*M+S+264.

The terminal enclosure follows from positive integral defects on the closed blocks of lengths 13 and 21 at orders 9 and 10, followed by the actual scalar child split through order 21. Put P(9)=13 and P(k)=floor((k-2)^2/3)-3*k+30 for 10<=k<=21. The zero child gaps lie on their proved platforms. The budget inequalities P(k)>=P(k-1)+platformWidth(k-2) and P(k)>=P(k-2)+platformWidth(k-1) yield v<=Q(k,v)*P(k) for positive defects. Applying this to the same terminal deficit-four root gives 348. The telescoping sum uses only outgoing orders at least 22.

There exist uniform nonnegative real constants a,c with b<=a*m+c for every legal actual deficit-four root of every order m>=21 if and only if there exist uniform nonnegative real constants a’,c’ with S<=a’M+c’ for every complete actual deficit-four spine starting at M>=22. Forward, S<=Phi(M,b0)<=b0 preserves a,c. Reverse, a=a’+4 and c=c’+264 give the root bound; the order-21 case is absorbed because 421+264=348. This equivalence does not establish the existence of either bound. No interval description of the full support, quadratic attainment or infinite descending spine is asserted.

References