1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0
Triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 ...
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[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2, 1,
[5] 1, 1, 4, 3, 3, 1,
[6] 1, 1, 5, 4, 6, 3, 1,
[7] 1, 1, 6, 5, 10, 6, 4, 1,
[8] 1, 1, 7, 6, 15, 10, 10, 4, 1,
[9] 1, 1, 8, 7, 21, 15, 20, 10, 5, 1,
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From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
N = 5: x^2 - 5;
N = 7: x^3 - 7x + 7;
N = 11: x^5 - 11x^3 + 11x^2 + 11x - 11;
N = 13: x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
N = 17: x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
N = 19: x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)
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