Half interval

• 2026-10-02 [Proved] At an odd base base = 2 fill - 1 the half interval F = {0..fill-1} has 2 S_F equal to the design on the even digits, and at a prime base p its set S_F is {k >= 1 : p does not divide C(2k,k)}, since Legendre's formula makes v_p(C(2k,k)) the number of carries in k + k. Witness: mobius The half interval, bullet The object.
• 2026-10-02 [Proved] For every digit set the shifted grid sums satisfy Sigma_i(s) = (T^i 1)(base^i s) with (T phi)(t) = sum_(r < base) abs(hat F((t+r)/base)) phi((t+r)/base), so any phi with 1 <= phi <= Phi and T phi <= lambda phi is a shifted-grid certificate with C_F = Phi and base^(alpha_1) = lambda/fill. Witness: mobius The half interval, bullet The transfer operator.
• 2026-10-02 [Proved] At every odd base >= 3 the half interval has T phi <= lambda phi for phi = 1 + abs(sin(pi t))/2 and lambda = fill + X_0 + csc(pi/(2 base))/2, X_0 = (base/pi)(log(base + 3) + gamma + log tan(3 pi/8 + pi/(4 base))) + (sqrt 2 - 4/pi)(base + 1)^2/(8 base), so Sigma_i(s) <= (3/2) lambda^i at every level and shift. Witness: mobius The half interval, bullet The certificate.
• 2026-10-02 [Proved] The per-digit l^1 cost of the half interval is (2/pi) log base + O(1) from both sides: lambda/fill <= (2/pi) log base + 2.60043004 + 3.4/base at every odd base >= 101, checked directly at odd 101..3001 and four larger bases, and at every odd base >= 9 every grid sum at level i is at least (((2/pi) log base - 1.31) fill)^i. Witness: mobius The half interval, bullet The constant is 2/pi; lab/py/interval-digits verb wall.
• 2026-10-02 [Proved] The half interval carries a shifted-grid certificate with alpha_1 < 1/5 at every odd base >= 94939, where 1/5 - alpha_1 >= 1.3678 * 10^-8, and with alpha_1 < 1/4 at every odd base >= 3789, where 1/4 - alpha_1 >= 7.9625 * 10^-6, each certified at 120 bits up to a monotone tail bound. Witness: lab/py/interval-digits verb wall; mobius The half interval, bullet The wall.
• 2026-10-02 [Proved] At every odd base >= 94939 the half interval has abs(M_F(x)) <= C A_F(x) exp(-c sqrt(log x)) and abs(sum_(n <= x, n in S_F) Lambda(n) - kappa_F A_F(x)) <= C A_F(x) exp(-c sqrt(log x)) at every x >= 2, kappa_F = (base/phi(base)) #{1 <= f < fill : gcd(f, base) = 1}/fill, with C, c > 0 computable from base alone. Witness: mobius The half interval, bullet The theorem on the half interval; coprime The half interval.
• 2026-10-02 [Proved] At every prime p >= 94939 the sum of mu(k) over k <= x with p not dividing C(2k,k) is at most C A(x) exp(-c sqrt(log x)) in absolute value, and the sum of Lambda(k) over the same k is p A(x)/(p+1) within C A(x) exp(-c sqrt(log x)), A(x) the count of such k. Witness: mobius The half interval, bullet The theorem on the half interval.
• 2026-10-02 [Proved] Under the generalized Riemann hypothesis for every Dirichlet character, at every odd base >= 3789 the half interval has abs(M_F(x)) <<_(base, eps) A_F(x)^(1 - delta + eps) with delta = (1/4 - alpha_1)/alpha_base > 0, and 1 - delta tends to 3/4 as the base grows. Witness: mobius The half interval, bullet The theorem on the half interval; coprime bullet What the bar 1/4 buys.
• 2026-10-02 [Proved] If for every eps > 0 some C_eps gives abs(M_F(x)) <= C_eps A_F(x)^(1/2 + eps) on the half interval at every odd base and every x >= 1, the Riemann hypothesis holds, trivially, since S_F contains every integer below fill; no converse is claimed. Witness: mobius The half interval, bullet Uniform square-root cancellation over the bases implies RH.
• 2026-10-02 [Verified] Maynard 2022 Theorem 1.3 at consecutive excluded digits with q - s >= q^(4/5 + eps), q large in terms of eps, contains the prime asymptotic on the half interval with a log-power error; it prints the main term with q - 1 where its 2019 constant has q - s, giving p/(2(p-1)) against p/(p+1) at a prime p, and its Section 9 constant clears 1/5 there only from q = 7777884825. Witness: arXiv:1510.07711v1 read at source; lab/py/interval-digits verb wall.
• 2026-10-02 [Verified] Maynard 2019 Theorem 1.2 gives the order of magnitude of the primes avoiding the top block B = {q - s..q-1} at s <= q - q^(57/80) for q large, a range holding the half interval, and its remark gives the asymptotic only for the primes avoiding the bottom block B = {0..s-1} at s <= q - q^(3/4 + delta); neither Maynard paper carries a Mobius sum. Witness: arXiv:1604.01041v2 read at source.
• 2026-10-02 [Verified] The three sums of the certificate's proof hold on grids of shifts at every odd base 3..401 and at 1001, 10001, 100001, and the bounds on G and T phi and the lower bound on G at every odd base 3..401 and at five larger bases 1001..100001, G reaching at most 0.999857 of its bound, and the 40-digit grid sums at nine bases 3..21 stay at most 0.557408 of (3/2) lambda^i. Witness: lab/py/interval-digits verb check.
• 2026-10-02 [Conjecture] The true growth rate of the half interval's transfer operator reads (2/pi) log base + 2.26 near base 7 * 10^4 and meets base^(1/5) between 70001 and 80001, and its one-step constant reads (2 sqrt2/pi) log base + 1.19, clearing 1/4 from 7075 and 1/5 from 317063. Witness: lab/py/interval-digits verb rate.
• 2026-10-03 [Proved] At every odd base >= 3 the half interval F = {0..fill-1}, base = 2 fill - 1, has kappa_F = (base/phi(base)) #{f in F : gcd(f, base) = 1}/fill = base/(base + 1), since the coprime residues pair as f and base - f, never equal at odd base, with exactly one of each pair at most fill - 1, so phi(base)/2 of them lie in F. Witness: mobius.md, The half interval.