# 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.