(E-a) d=4: combination {1: Fraction(7, 1), 2: Fraction(12, 1), 3: Fraction(16, 1), 4: Fraction(8, 1)} >= c_4 coefficients {1: 4, 2: 12, 3: 16, 4: 8} ; value 62/3
(E-a) d=5: combination {1: Fraction(12, 1), 2: Fraction(20, 1), 3: Fraction(40, 1), 4: Fraction(40, 1), 5: Fraction(16, 1)} >= c_5 coefficients {1: 5, 2: 20, 3: 40, 4: 40, 5: 16} ; value 64
(E-a) ABBB d=3: combination {1: Fraction(3, 1), 2: Fraction(6, 1), 3: Fraction(4, 1)} = c_3 coefficients {1: 3, 2: 6, 3: 4} ; value 11/2
      trivial bounds: 80/3 = 80/3 , 242/3 = 242/3 , 13/2 = 13/2
(E-b) averaged form of Lemma 4 for k = 1, 2, 3 re-derived by enumerating D_k: OK
(E-c) LP value d=4: 20.666666667  (claimed 62/3 = 20.666666667)
(E-c) LP value d=5: 64.000000000  (claimed 64 = 64.000000000)
(E-c) LP value d=6: 198.666666667  (claimed 596/3 = 198.666666667)
(E-c) LP value ABBB d=3 (C1=1/2, C2=8/5, no 3-space row): 5.500000000 (claimed 11/2)
(E-c) exact primal point D = (0, 5/9, 2/3, 5/12) is feasible with value 62/3: the d=4 LP optimum is exactly 62/3
(E-d) Lemma 4 (k = 1, 2, 3; C_1 = 2/3, C_2 = 2, C_3 = 6) holds on random 4D grids: 4400 instances tested: OK
(E-e) stacked grids: c_4(ABB) = 18 , c_5(ABB) = 54 , c_3(ABBB) = 24/5
(A2) n=4: lattice colouring in the n x n array, no wraparound: between 4 and 4 occurrences; (8/5)(n-6)(n-10) = 96/5; (8/5) n^2 = 128/5
(A2) n=5: lattice colouring in the n x n array, no wraparound: between 11 and 12 occurrences; (8/5)(n-6)(n-10) = 8; (8/5) n^2 = 40
(A2) n=9: lattice colouring in the n x n array, no wraparound: between 72 and 72 occurrences; (8/5)(n-6)(n-10) = -24/5; (8/5) n^2 = 648/5
(A2) n=10: lattice colouring in the n x n array, no wraparound: between 95 and 96 occurrences; (8/5)(n-6)(n-10) = 0; (8/5) n^2 = 160
(A2) n=11: lattice colouring in the n x n array, no wraparound: between 121 and 124 occurrences; (8/5)(n-6)(n-10) = 8; (8/5) n^2 = 968/5
(A2) n=20: lattice colouring in the n x n array, no wraparound: between 503 and 504 occurrences; (8/5)(n-6)(n-10) = 224; (8/5) n^2 = 640
(A2) n=40: lattice colouring in the n x n array, no wraparound: between 2279 and 2280 occurrences; (8/5)(n-6)(n-10) = 1632; (8/5) n^2 = 2560
(A2) n=60: lattice colouring in the n x n array, no wraparound: between 5335 and 5336 occurrences; (8/5)(n-6)(n-10) = 4320; (8/5) n^2 = 5760
(A2) (8/5)(n-6)(n-10) <= count(lattice colouring, any offset) <= (8/5) n^2 for 10 <= n <= 60: OK
     formula at n = 1..5: ['72', '256/5', '168/5', '96/5', '8'] -> the restriction n >= 6 (the claim says n >= 10) is needed
(A2) exact M(4) = 7 <= (8/5)*16 = 128/5
(D-box) n=3: stacked colouring of the n^3 array: 0..25 occurrences; 6(n-4)^2(n-6) = -18; 6 n^3 = 162
(D-box) n=6: stacked colouring of the n^3 array: 392..588 occurrences; 6(n-4)^2(n-6) = 0; 6 n^3 = 1296
(D-box) n=9: stacked colouring of the n^3 array: 2116..2645 occurrences; 6(n-4)^2(n-6) = 450; 6 n^3 = 4374
(D-box) n=12: stacked colouring of the n^3 array: 6144..7168 occurrences; 6(n-4)^2(n-6) = 2304; 6 n^3 = 10368
(D-box) n=16: stacked colouring of the n^3 array: 17424..19360 occurrences; 6(n-4)^2(n-6) = 8640; 6 n^3 = 24576
(D-box) 6(n-4)^2(n-6) <= count(stacked colouring) <= 6 n^3 for 6 <= n <= 16: OK
(S7) ABB, d=2, 3x3 window, quadratic potential of Section 7: max of Psi0 + Q over 512 patterns = 2 (attained by 72); class sums zero: OK
ALL CHECKS IN THIS FILE PASSED
