check_note.py  (Python 3.13.5; nmax = 4096)
[1] exact critical exponent: ce() against a direct loop and against the definition of the source
    600 random words: 0 differences from the loop; 300 short words: 0 differences from the definition; ce(schematische) = 3/2 (the source's exact 12/8-power), ce(nennen) = 2
    (0.0 s)
[2] Proposition 3.1 (universal reading): ce(x) = k + r/q >= alpha > k + r/N = ce(sigma(x))
    alpha = 101/100  q =   2, r =   1, N = 101: ce(x) = 3/2       ce(sigma(x)) = 102/101
    alpha = 11/10    q =   2, r =   1, N =  11: ce(x) = 3/2       ce(sigma(x)) = 12/11
    alpha = 5/4      q =   2, r =   1, N =   5: ce(x) = 3/2       ce(sigma(x)) = 6/5
    alpha = 4/3      q =   2, r =   1, N =   4: ce(x) = 3/2       ce(sigma(x)) = 5/4
    alpha = 3/2      q =   3, r =   2, N =   5: ce(x) = 5/3       ce(sigma(x)) = 7/5
    alpha = 5/3      q =   4, r =   3, N =   5: ce(x) = 7/4       ce(sigma(x)) = 8/5
    alpha = 199/100  q = 101, r = 100, N = 102: ce(x) = 201/101   ce(sigma(x)) = 101/51
    alpha = 201/100  q =   2, r =   1, N = 101: ce(x) = 5/2       ce(sigma(x)) = 203/101
    alpha = 7/3      q =   2, r =   1, N =   4: ce(x) = 5/2       ce(sigma(x)) = 9/4
    alpha = 5/2      q =   3, r =   2, N =   5: ce(x) = 8/3       ce(sigma(x)) = 12/5
    alpha = 299/100  q = 101, r = 100, N = 102: ce(x) = 302/101   ce(sigma(x)) = 152/51
    alpha = 17/5     q =   2, r =   1, N =   3: ce(x) = 7/2       ce(sigma(x)) = 10/3
    alpha = 1001/100 q =   2, r =   1, N = 101: ce(x) = 21/2      ce(sigma(x)) = 1011/101
    alpha = 41/4     q =   2, r =   1, N =   5: ce(x) = 21/2      ce(sigma(x)) = 51/5
    (0.0 s)
[3] Proposition 4.1: one letter; repeated letters and 1 + 1/s; binary words never have ce in (3/2, 2)
    s = 2: all 16382 words of length <= 13: a repeated letter <=> ce >= 1 + 1/s <=> ce > 1
    s = 3: all 29523 words of length <= 9: a repeated letter <=> ce >= 1 + 1/s <=> ce > 1
    s = 4: all 21844 words of length <= 7: a repeated letter <=> ce >= 1 + 1/s <=> ce > 1
    s = 5: all 19530 words of length <= 6: a repeated letter <=> ce >= 1 + 1/s <=> ce > 1
    all 32766 binary words of length <= 14: ce is never in (3/2, 2); square-free ones have length <= 3
    (0.4 s)
[4] Lemma 4.2 (Thue-Morse desubstitution) on examples
    ce(mu^n(0)) = 2 for 2 <= n <= 14; 459 periods p with a factor longer than 2p in mu^n(w) (150 random w, n <= 3): 2^n | p and w has period p/2^n on a factor with at least the same ratio
    (0.2 s)
[5] Proposition 4.3: x = (01)^k 0, sigma_n(1) = mu^n(0) without its first letter
    k = 2..5, n = 1..8: ce(x) = k + 1/2 and ce(sigma_n(x)) = k + 2^-n in every case
    (0.0 s)
[6] Proposition 4.4: three letters, x = (Tc)^k T, sigma(c) = cU
    k = 2..4, j = 0..3, r = 2, 4: ce(x) = k + 2^j/(2^j+1) and ce(sigma(x)) = k + 2^j/(2^(j+r)+1)
    (0.0 s)
[7] Proposition 4.5: certificates for r < q < N <= 9 and the union of the intervals for s <= 30
    exact values ce(x) = 1 + r/q, ce(sigma(x)) = 1 + r/N; union = (4/3, 3/2] for s = 3 and (1 + 1/s, 1 + (s-2)/(s-1)] for 4 <= s <= 30
    (0.2 s)
[8] Lemma 5.1 for the family F, 2 <= N <= 1536: letters, sums, lengths, squares of E^Z
    20 words: e_t in {0,1,2}, e_1 + ... + e_N = N, |tau(E)| = 2N, |M(E)| = 12N, and E^Z has no square uu with N not dividing |u|
    (0.0 s)
[9] Lemma 5.2: occurrences of B in B A^a B A^b B, all nine pairs (a, b)
    B = 010010011, A = 001: B occurs only at the three positions 0, 9 + 3a, 18 + 3a + 3b
    (0.0 s)
[10] Lemma 5.3: all square-free ternary words of length 14, all periods p < 42
    456 windows B A^(w_1) B ... A^(w_14) B: the largest ratio |f|/p over p < 42 is 8/3 (p = 3)
    (0.1 s)
[11] Lemma 5.4 (N in {1, 2, 3, 4, 6}) and a cross-check of Lemma 5.5 (8 <= N <= 192): factors of M^Z
    M(E) primitive, sup of |f|/p over periods p not divisible by |M| (exact, cyclic):
    (2):1:8/3, G_2:2:8/3, H_1:3:8/3, G_3:4:8/3, H_2:6:8/3, G_4:8:8/3, H_3:12:8/3, G_5:16:8/3, H_4:24:8/3, G_6:32:8/3, H_5:48:8/3, G_7:64:8/3, H_6:96:8/3, G_8:128:8/3, H_7:192:8/3
    (0.0 s)
[12] constants in the proof of Lemma 5.5
    floor(8*41/3) + 1 = 110; 109 <= 13*9 + 8; gaps 9 + 3e <= 15; 2p + 28 <= 8p/3 exactly for p >= 42
    (0.0 s)
[13] Proposition 5.6: exact ce for N <= 24 and k = 2, 3, 4; the certificate K2; covering of (k, k+1)
    ce(x) = k + 1 - 1/|tau(E)|, ce(sigma(x)) = max(8/3, k + 1 - 9/|M(E)|) for all 27 pairs (N <= 24); K2: ce(00100100) = 8/3, ce(001011010010110100) = 9/4;
    for k = 2..6 the intervals with n <= 30 and N <= 3072 form the single interval (k + 2^-30, k + 1 - 1/6144]
    (0.0 s)
[14] Proposition 6.1: the certificates (a) and the certificates (b) for N <= 4096
    N =      4: |x_N| =      7, |sigma0(x_N)| =     13, ce(x_N) = 7/4           ce(sigma0(x_N)) = 5/3           (0.0 s)
    N =      6: |x_N| =     11, |sigma0(x_N)| =     21, ce(x_N) = 11/6          ce(sigma0(x_N)) = 7/4           (0.0 s)
    N =      8: |x_N| =     15, |sigma0(x_N)| =     29, ce(x_N) = 15/8          ce(sigma0(x_N)) = 11/6          (0.0 s)
    N =     12: |x_N| =     23, |sigma0(x_N)| =     45, ce(x_N) = 23/12         ce(sigma0(x_N)) = 15/8          (0.0 s)
    N =     16: |x_N| =     31, |sigma0(x_N)| =     61, ce(x_N) = 31/16         ce(sigma0(x_N)) = 23/12         (0.0 s)
    N =     24: |x_N| =     47, |sigma0(x_N)| =     93, ce(x_N) = 47/24         ce(sigma0(x_N)) = 31/16         (0.0 s)
    N =     32: |x_N| =     63, |sigma0(x_N)| =    125, ce(x_N) = 63/32         ce(sigma0(x_N)) = 47/24         (0.0 s)
    N =     48: |x_N| =     95, |sigma0(x_N)| =    189, ce(x_N) = 95/48         ce(sigma0(x_N)) = 63/32         (0.0 s)
    N =     64: |x_N| =    127, |sigma0(x_N)| =    253, ce(x_N) = 127/64        ce(sigma0(x_N)) = 95/48         (0.0 s)
    N =     96: |x_N| =    191, |sigma0(x_N)| =    381, ce(x_N) = 191/96        ce(sigma0(x_N)) = 127/64        (0.0 s)
    N =    128: |x_N| =    255, |sigma0(x_N)| =    509, ce(x_N) = 255/128       ce(sigma0(x_N)) = 191/96        (0.0 s)
    N =    192: |x_N| =    383, |sigma0(x_N)| =    765, ce(x_N) = 383/192       ce(sigma0(x_N)) = 255/128       (0.0 s)
    N =    256: |x_N| =    511, |sigma0(x_N)| =   1021, ce(x_N) = 511/256       ce(sigma0(x_N)) = 383/192       (0.0 s)
    N =    384: |x_N| =    767, |sigma0(x_N)| =   1533, ce(x_N) = 767/384       ce(sigma0(x_N)) = 511/256       (0.0 s)
    N =    512: |x_N| =   1023, |sigma0(x_N)| =   2045, ce(x_N) = 1023/512      ce(sigma0(x_N)) = 767/384       (0.0 s)
    N =    768: |x_N| =   1535, |sigma0(x_N)| =   3069, ce(x_N) = 1535/768      ce(sigma0(x_N)) = 1023/512      (0.0 s)
    N =   1024: |x_N| =   2047, |sigma0(x_N)| =   4093, ce(x_N) = 2047/1024     ce(sigma0(x_N)) = 1535/768      (0.0 s)
    N =   1536: |x_N| =   3071, |sigma0(x_N)| =   6141, ce(x_N) = 3071/1536     ce(sigma0(x_N)) = 2047/1024     (0.0 s)
    N =   2048: |x_N| =   4095, |sigma0(x_N)| =   8189, ce(x_N) = 4095/2048     ce(sigma0(x_N)) = 3071/1536     (0.0 s)
    N =   3072: |x_N| =   6143, |sigma0(x_N)| =  12285, ce(x_N) = 6143/3072     ce(sigma0(x_N)) = 4095/2048     (0.1 s)
    N =   4096: |x_N| =   8191, |sigma0(x_N)| =  16381, ce(x_N) = 8191/4096     ce(sigma0(x_N)) = 6143/3072     (0.2 s)
    the intervals (ce(sigma0(x_N)), ce(x_N)] chain; their union is (5/3, 8191/4096]
    (0.4 s)
[15] Remark 6.2: sigma0 = pi o theta, theta(E_N) = E_2N and theta(x_N) = (E_2N E_2N without 3 letters)
    letterwise identity, theta(gap word of z) = gap word of mu(z) for z = mu^n(0) (n <= 11), mu^n(001) (n <= 10),
    and theta(x_N) = (E_2N E_2N)[3:] for N <= 1536
    (0.0 s)
TOTAL failures: 0   (running time 1 s)
