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EXAMPLE
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The irregular triangle T, with entries T(n, m), begins: (increasing k >= 1 values are separated by ;)
n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
1: [1, 0]
2: [1, 1]
3: [2, 1], [1, 2]; [1, 0, 1]
4: [3, 1], [1, 3]; [2, 0, 1], [1, 0, 2], [1, 1, 1]
...
n = 5: [4, 1], [1, 4], [3, 2], [2, 3]; [3, 0, 1], [1, 0, 3], [2, 1, 1], [1, 2, 1], [1, 1, 2]; [2, 0, 0, 1], [1, 0, 0, 2], [1, 1, 0, 1], [1, 0, 1, 1]; [1, 0, 0, 0, 1]
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n = 6: [5, 1], [1, 5]; [4, 0, 1], [1, 0, 4], [3, 0, 2], [2, 0, 3], [3, 1, 1], [1, 3, 1], [1, 1, 3], [2, 2, 1], [2, 1, 2], [1, 2, 2]; [3, 0, 0, 1], [1, 0, 0, 3], [2, 1, 0, 1], [2, 0, 1, 1], [1, 2, 0, 1], [1, 0, 2, 1], [1, 1, 0, 2], [1, 0, 1, 2], [1, 1, 1, 1]; [2, 0, 0, 0, 1], [1, 0, 0, 0, 2], [1, 1, 0, 0, 1], [1, 0, 1, 0, 1], [1, 0, 0, 1, 1]
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n = 7: [6, 1], [1, 6], [5, 2], [2, 5], [4, 3], [3, 4]; [5, 0, 1], [1, 0, 5], [4, 1, 1], [1, 4, 1], [1, 1, 4], [3, 2, 1], [3, 1, 2], [2, 3, 1], [2, 1, 3], [1, 3, 2], [1, 2, 3]; [4, 0, 0, 1], [1, 0, 0, 4], [3, 0, 0, 2], [2, 0, 0, 3], [3, 1, 0, 1], [3, 0, 1, 1], [1, 3, 0, 1], [1, 0, 3, 1], [1, 1, 0, 3], [1, 0, 1, 3], [2, 2, 0, 1], [2, 0, 2, 1], [2, 1, 0, 2], [2, 0, 1, 2], [1, 2, 0, 2], [1, 0, 2, 2], [2, 1, 1, 1], [1, 2, 1, 1], [1, 1, 2, 1], [1, 1, 1, 2]; [3, 0, 0, 0, 1], [1, 0, 0, 0, 3], [2, 1, 0, 0, 1], [2, 0, 1, 0, 1], [2, 0, 0, 1, 1], [1, 2, 0, 0, 1], [1, 0, 2, 0, 1], [1, 0, 0, 2, 1], [1, 1, 0, 0, 2], [1, 0, 1, 0, 2], [1, 0, 0, 1, 2], [1, 1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 0, 1, 1, 1]; [2, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 2], [1, 1, 0, 0, 0, 1], [1, 0, 1, 0, 0, 1], [1, 0, 0, 1, 0, 1], [1, 0, 0, 0, 1, 1]
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x^6 + 1 = (x^2 + 1)*(x^4 - x^2 + 1), hence no [1, 0, 0, 0, 0, 0, 1] recorded, because x^6 - 1 also factorizes.
...
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Polynomials: n = 4, degree k = 1: 3*x + 1, x + 3; k = 2: 2*x^2 + 1, x^2 + 2, x^2 + x + 1; k = 3: no entry [1, 0, 0, 1], because x^3 + 1 factorizes, as well as x^3 - 1.
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Height n = 4, degree k = 2, with signed polynomials:
[2, 0, 1] for 2*x^2 + 1, 2*x^2 - 1, [1, 0, 2] for x^2 + 2, x^2 - 2, and [1, 1, 1] for x^2 + x + 1, x^2 + x - 1, x^2 - x + 1, x^2 - x - 1. The corresponding real algebraic numbers come in signed pairs only from 2*x^2 - 1, x^2 - 2, x^2 + x - 1, and x^2 - x - 1, namely, -sqrt(1/2), +sqrt(1/2), -sqrt(2), +sqrt(2), -phi = -A001622, phi - 1, and -(phi - 1), phi. So Cantor's phi (our Phi) is Phi(4, 2) = 8. Together with the four real k = 1 roots from the signed polynomials for [3, 1] and [1, 3] one finds Phi(4) = 12. See A362364.
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