Digit Polygons from Repeating Expansions: Exact Area Identities, Finite-Lag Reduction, and Base-b Asymptotics

Let b >= 2 be an integer and p a prime with gcd(b,p)=1. Write the base‑b expansion of 1/p as 0.d0 d1 ... dₖ₋₁ (repeating), where k = ord_p(b) is the multiplicative order of b modulo p. The digit polygon is the closed polygon with vertices (d_i, dᵢ₊₁) for i=0,...,k-1 (indices mod k). This work isolates the mathematically robust core of the digit‑polygon construction and adds a reverse construction. The reverse construction starts from a pointed directed polygon on the digit‑pair grid {0,1,…,b-1}^2. Such a polygon determines a repeating base‑b expansion exactly when consecutive vertices overlap, i.e., when V_i = (x_i,y_i) satisfies y_i = xᵢ₊₁ cyclically. The resulting digit block d0…dₖ₋₁ yields the repeating expansion 0.(d0 d1 … dₖ₋₁) = N/(b^k - 1) with N = d0 bᵏ⁻¹ + d1 bᵏ⁻² + … + dₖ₋₁. Prime reciprocals are an arithmetic subfamily: the reverse‑generated expansion equals 1/p for a prime p exactly when N divides b^k - 1 and (b^k - 1)/N is prime. The signed area A_b(p) satisfies an exact Dirichlet‑energy identity: A_b(p) = -1/4 * sum_i (dᵢ₊₂ - d_i)^2. This immediately implies A_b(p) < 0 for every prime with ord_p(b) >= 3. The Fourier representation is A_b(p) = -1/(k p^2) * sum_j |b - ω^j|^2 sin^2(2π j/k) |r̂_j|^2, where r_i = b^i mod p and ω = e2π i/k. Expanding in residues yields a finite‑lag reduction: the area depends on exactly four orbit correlations, with coefficient polynomial P_b(z) = -(b^2+1) + b z + (b^2+1)z^2 - b z^3 = -(z^2-1)(b z - (b^2+1)). For two digit sequences of the same period we prove an exact discriminant theorem: the Gram determinant of the area form equals a Cauchy‑Schwarz determinant for the lag‑2 difference operator; it is always nonnegative, and equality is characterized completely by the kernel of that operator. Finally, the reverse construction gives a random‑digit baseline. For independent uniform base‑b digits, the expected normalized area is -Var(U_b)/2 = -(b^2 - 1)/24, where U_b is uniform on {0,…,b-1}. For full‑reptend primes (ord_p(b) = p-1) we prove the matching asymptotic law: A_b(p)/(p-1) = -(b^2-1)/24 + O_b(1/p).