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EXAMPLE
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Given C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the Catalan function of A000108, which starts C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + ... + binomial(2*n,n)/(n+1) * x^n + ...
then
1+x - x^2*C(x^3) = 1 + x - x^2 - x^5 - 2*x^8 - 5*x^11 - 14*x^14 - 42*x^17 - 132*x^20 - 429*x^23 - 1430*x^26 - ...
and the table of coefficients of x^k in (1+x - x^2*C(x^3))^(n*(2*n+1)) begin:
n=1: [1, 3, 0, -5, 0, 0, -7, 3, ...];
n=2: [1, 10, 35, 30, -105, -238, 0, 270, ...];
n=3: [1, 21, 189, 910, 2205, 378, -13321, -33324, ...];
n=4: [1, 36, 594, 5880, 38115, 162756, 407862, 175068, ...];
n=5: [1, 55, 1430, 23265, 263835, 2193191, 13612995, 62337330, ...];
n=6: [1, 78, 2925, 70070, 1201200, 15633540, 159772613, 1305975528, ...];
...
from which the terms a(n) = [x^(n+1)] (1+x - x^2*C(x^3))^(n*(2*n+1))/(n*(2*n+1)) can be derived, for n >= 1, as illustrated by:
a(1) = [x^2] (1+x - x^2*C(x^3))^3 / 3 = 0 / 3 = 0;
a(2) = [x^3] (1+x - x^2*C(x^3))^10 / 10 = 30 / 10 = 3;
a(3) = [x^4] (1+x - x^2*C(x^3))^21 / 21 = 2205 / 21 = 105;
a(4) = [x^5] (1+x - x^2*C(x^3))^36 / 36 = 162756 / 36 = 4521;
a(5) = [x^6] (1+x - x^2*C(x^3))^55 / 55 = 13612995 / 55 = 247509;
a(6) = [x^7] (1+x - x^2*C(x^3))^78 / 78 = 1305975528 / 78 = 16743276;
...
Apparently, all terms computed this way are divisible by 3 (verified for the initial 1201 terms).
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