== A. rates of atomic Xi: DP versus brute force; consistency; (5), (6), closed forms
PASS DP rates = brute force, b<=6, Xi=(Fraction(0, 1), [(Fraction(1, 1), [Fraction(1, 4), Fraction(1, 8)])]) | mismatches 0
PASS Xi rates satisfy the consistency relation (Dong (9)), levels <= 8 | violations 0
PASS (5) T(b+1)-T(b) = b t_{b+1} and (6) t_{b-1} = t_b + tau_b + (b-3) j_b, b<=8
PASS closed forms of t_n, j_n, T(n) agree with the type sums, n<=8
PASS DP rates = brute force, b<=6, Xi=(Fraction(1, 3), [(Fraction(2, 1), [Fraction(1, 2), Fraction(1, 3)]), (Fraction(1, 1), [Fraction(3, 5)])]) | mismatches 0
PASS Xi rates satisfy the consistency relation (Dong (9)), levels <= 8 | violations 0
PASS (5) T(b+1)-T(b) = b t_{b+1} and (6) t_{b-1} = t_b + tau_b + (b-3) j_b, b<=8
PASS closed forms of t_n, j_n, T(n) agree with the type sums, n<=8
PASS DP rates = brute force, b<=6, Xi=(Fraction(0, 1), [(Fraction(1, 1), [Fraction(1, 3), Fraction(1, 3), Fraction(1, 3)])]) | mismatches 0
PASS Xi rates satisfy the consistency relation (Dong (9)), levels <= 8 | violations 0
PASS (5) T(b+1)-T(b) = b t_{b+1} and (6) t_{b-1} = t_b + tau_b + (b-3) j_b, b<=8
PASS closed forms of t_n, j_n, T(n) agree with the type sums, n<=8
PASS DP rates = brute force, b<=6, Xi=(Fraction(0, 1), [(Fraction(1, 1), [Fraction(2, 5), Fraction(1, 5), Fraction(1, 10)])]) | mismatches 0
PASS Xi rates satisfy the consistency relation (Dong (9)), levels <= 8 | violations 0
PASS (5) T(b+1)-T(b) = b t_{b+1} and (6) t_{b-1} = t_b + tau_b + (b-3) j_b, b<=8
PASS closed forms of t_n, j_n, T(n) agree with the type sums, n<=8
== B. EPPF: Ewens for Kingman; consistency; total mass; Kingman and star regenerative
PASS Kingman a=1 rho=1/2: EPPF = Ewens(theta=2rho/a), n<=7 | mismatches 0
PASS Kingman: (R_n) holds at every partition, n<=7
PASS Kingman a=3/2 rho=2: EPPF = Ewens(theta=2rho/a), n<=7 | mismatches 0
PASS Kingman: (R_n) holds at every partition, n<=7
PASS star c=1 rho=1/3: (R_n) holds at every partition, n<=7
PASS star c=2 rho=5/7: (R_n) holds at every partition, n<=7
PASS EPPF addition rule (exchangeable, consistent) n<=6 and total mass 1, n<=7
== C. Lemma 2.2 (hook q >= 0; full system), Lemma 2.3, Lemma 3.1 on random inputs
   trial 0 (arbitrary consistent array, N=8, rho=1/5) done
   trial 1 (atomic Xi, N=8, rho=2) done
   trial 2 (arbitrary consistent array, N=8, rho=2/3) done
   trial 3 (atomic Xi, N=8, rho=4) done
   trial 4 (arbitrary consistent array, N=7, rho=9/2) done
   trial 5 (atomic Xi, N=7, rho=2) done
   trial 6 (arbitrary consistent array, N=7, rho=4) done
   trial 7 (atomic Xi, N=7, rho=1) done
PASS Lemma 2.2: hook-determined q(n;m) >= 0, all (trial, n) | 52/52
PASS Lemma 2.2: full system solvable iff non-hook residuals vanish; then q = hook q, sum 1 | 40/40
PASS Lemma 2.3: H_m = T(m)/(rho m(m-1)) | 52/52
PASS Lemma 3.1: eps_n = p(1^n) Delta_n/(rho^2 C(n,2)^2 C(n-2,2)), 4<=n<=8 | 36/36
PASS negative control: W(n) in place of W(n-1) is detected at n=5,6,7
== E. Theorem A (rho*), Corollary 3.2 (Lambda case)
PASS Theorem A: Pi_4 regenerative exactly at rho = rho*(Xi) (non-Lambda atomic Xi) | 27/27
PASS Corollary 3.2: Lambda-coalescents regenerative at level 4 iff Kingman or star | 6/6
== F. Proposition 7.1: Xi = (17/1024) delta_0 + delta_(1/4,1/8), rho = 13833/5120
PASS rho = rho*(Xi) | rho* = 13833/5120
PASS level 2: full system has a unique solution, q >= 0, sum 1 | q = (9222/10957, 1735/10957)
PASS level 3: full system has a unique solution, q >= 0, sum 1 | q = (53/70, 10229/54785, 43063/766990)
PASS level 4: full system has a unique solution, q >= 0, sum 1 | q = (106/149, 53/298, 421721/4897779, 241301/9795558)
PASS level 5: full system has a unique solution, q >= 0, sum 1 | q = (23055/33517, 5300/33517, 9805/100551, 16183550/367245769, 1245097/100157937)
PASS level 6: full system has no solution | hook-residual nonzero at 5 partitions: [(4, 2), (3, 3), (3, 2, 1), (2, 2, 2), (2, 2, 1, 1)]
PASS level 7: full system has no solution | hook-residual nonzero at 8 partitions: 
PASS q(5;.) equals the vector printed in Proposition 7.1
PASS level 6 fails exactly at (4,2),(3,3),(3,2,1),(2,2,2),(2,2,1,1)
   residual at (4, 2): -1072617989416/62545017657659229
   residual at (3, 3): 3435702810172/187635052972977687
   residual at (3, 2, 1): 39121971094/62545017657659229
   residual at (2, 2, 2): -92094918940/62545017657659229
   residual at (2, 2, 1, 1): 367050883928/62545017657659229
PASS closed-form Delta_n equals Delta_n from the type sums, n<=7
   sign of Delta_n, n=4..60: 00++++++++++---------------------------------------------
PASS Delta_4 = Delta_5 = 0, Delta_n > 0 for 6<=n<=15, Delta_n < 0 for 16<=n<=60
   Bin(60)W(59)/C(60,2) = 0.2124999993   (aM = 17/80 = 0.2125000000)
PASS Bin(n)W(n-1)/C(n,2) close to aM = 17/80 at n = 60
== G. Theorem C example; eventual signs of Delta_n (Theorems B and D)
PASS Xi = delta_(1/3,1/3,1/3): Bin(5) = 0, X(5) = 10/9
PASS   ... and (R_5) fails (Delta_5 > 0)
PASS Theorem C, Xi = delta_(1/2,1/2) + 2 delta_(1/4,1/4,1/4,1/4) (K = 4): Bin(6) = 0, X(6) > 0, Delta_6 > 0
PASS Theorem C remark: K = 2 gives t_4 = 0 (so Pi_4 fails by Theorem A)
PASS Theorem D, K*=2: Delta_n has sign + for 40<=n<=60 (rho=1)
PASS Theorem D, K*=3: Delta_n has sign + for 40<=n<=60 (rho=1)
PASS Theorem D, K*=1: Delta_n has sign - for 40<=n<=60 (rho=1)
PASS Theorem D, K*=1 vs dust-free: Delta_n has sign - for 40<=n<=60 (rho=1)
PASS Theorem B: Delta_n has sign - for 40<=n<=60 (rho=1)
PASS Theorem B: Delta_n has sign - for 40<=n<=60 (rho=1)
== H. Lemmas 4.1, 4.2, (Q1) exactly; Lemma 4.3 constant c(1/4)
PASS Lemma 4.3 constants: theta in (0,1/2] meets both conditions; c(1/4) > 0 | s0=0.04868 p0=1.853e-05 theta=7.863e-03 c(1/4)=1.519e-08
PASS Lemma 4.1 pi_2 <= lambda_m^2/2 (40 paintboxes, 3<=m<=30, exact) | violations 0
PASS Lemma 4.2 pi_2(m) <= m/(m-4) pi_2(m-1), m>=5 | violations 0
PASS (Q1) F(x;m), G(x;m) nondecreasing in m | violations 0
PASS Lemma 4.3 at eta=1/4: G(x;m) >= c(1/4) whenever lambda_m >= 1/4 | cases 980, smallest G seen 0.02878
TOTAL: 57 checks, 57 passed, 0 failed
