Part 1. Ground truth
  ok  recursion b_{2n+1}=b_n, b_{4n}=b_n, b_{4n+2}=0 == block definition (n < 2**14)         [   0.0 s]
      Euclid: 23746 quotients for alpha_0, 44641 for {f}
  ok  det(b_{i+j})_{i,j<n} != 0  <=>  n in Sigma_0 (Euclid on alpha_0), n <= 300             [   2.4 s]
  ok  det(b_{i+j+1})_{i,j<n} != 0  <=>  n in Sigma_1 (Euclid on {f}), n <= 300               [   4.0 s]
Part 2. The Mills-Robbins expansion
  ok  {f} = [0; X, Gamma] : Gamma == quotients of {f} by Euclid (44640 letters)              [   4.3 s]
  ok  Lemma 3.1: D_n = 2**(n+1) - 1 (n <= 19)                                                [   4.3 s]
  ok  Lambda_n palindromic, of total degree 2**(n-1) + 2                                     [   4.3 s]
  ok  phi_1: |G_m| = 2**m, A_m = A_{m-1} G_m, B_m = G_m B_{m-1} (5 <= m <= 19)               [   4.5 s]
  ok  phi_0: |G_m| = 2**m, A_m = A_{m-1} G_m, B_m = G_m B_{m-1} (5 <= m <= 19)               [   4.8 s]
  ok  chi_1 = 11 . phi_1(Gamma)  (dictionary for {f}; n < 65536)                             [   4.8 s]
Part 3. Factors of length 3
  ok  |F| = 20 and F closed under reversal                                                   [   4.8 s]
  ok  base case: Fac_3(Gamma_n) subset F for 3 <= n <= 8                                     [   4.8 s]
  ok  Fac_3(Lambda_n) subset F (n <= 18; constant in n for n >= 5)                           [   4.9 s]
  ok  begin_2(Gamma_n) constant, end_3(Gamma_n) depends on n mod 3                           [   4.9 s]
  ok  induction step: the 7 boundary factors of each class n mod 6 lie in F                  [   4.9 s]
  ok  no two adjacent letters of degree 2                                                    [   4.9 s]
Part 4. The transducer for t/f
  ok  prefix: t/f = [X+1; X, ...], nothing emitted at Gamma[1] = X**2+1                      [   4.9 s]
  ok  after the prefix, (output degree E) - (degree D read in Gamma) = -1                    [   4.9 s]
  ok  product transducer x F: 12 configurations, determinant = t in all                      [   4.9 s]
  ok  output alphabet {X, X**3, X**3+X}; every emission certain for all admissible tails     [   4.9 s]
  ok  a degree-1 letter emits one quotient (X, X**3, X**3+X after deg 1, X**2, X**2+1), then E - D = 1; a degree-2 letter emits nothing, then E - D = -1 [   4.9 s]
  ok  hence chi_0 = 11 . phi_0(Gamma); checked against Euclid (n < 65536)                    [   4.9 s]
  ok  transfer relation chi_0(n) = chi_1(n) chi_1(n-1)                                       [   5.0 s]
Part 5. The automaton
  ok  e=1: base table n < 256 and leading-zero invariance                                    [   5.0 s]
  ok  e=1: rules (R1) omega(111y) = omega(y), (R2) omega(100 1..1 0 z) = omega(1z), (R3) B_4 block [   5.0 s]
  ok  e=1: rule (R4): the block L_m (both parities, all positions)                           [   5.0 s]
  ok  e=1: automaton == chi_1(n+1) on the whole prefix (n < 1048575)                         [   6.3 s]
  ok  e=1: automaton == Euclid's Sigma_1 directly (1 <= n < 65536)                           [   6.3 s]
  ok  e=0: base table n < 256 and leading-zero invariance                                    [   6.4 s]
  ok  e=0: rules (R1) omega(111y) = omega(y), (R2) omega(100 1..1 0 z) = omega(1z), (R3) B_4 block [   6.4 s]
  ok  e=0: rule (R4): the block L_m (both parities, all positions)                           [   6.4 s]
  ok  e=0: automaton == chi_0(n+1) on the whole prefix (n < 1048575)                         [   7.8 s]
  ok  e=0: automaton == Euclid's Sigma_0 directly (1 <= n < 65536)                           [   7.8 s]
Part 6. Kernels and minimal automata of chi
  ok  chi_0: minimal MSB-first automaton has 15 states                                       [   8.0 s]
  ok  chi_0: 2-kernel has 74 elements (= states of the minimal LSB-first automaton)          [   8.0 s]
  ok  chi_1: minimal MSB-first automaton has 8 states                                        [   8.1 s]
  ok  chi_1: 2-kernel has 68 elements (= states of the minimal LSB-first automaton)          [   8.1 s]
Part 7. Linear complexity profile
  ok  Berlekamp-Massey L(b,N) == deg Q_j with deg Q_{j-1}+deg Q_j <= N < deg Q_j+deg Q_{j+1} (N <= 3000) [   8.2 s]
  ok  gaps of Sigma_0 are 1 and 3 (up to 65536)                                              [   8.2 s]
  ok  2L(b,N) - N takes exactly the values -2..3 (N < 65534)                                 [  54.1 s]
ALL CHECKS PASSED