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%I #21 Sep 30 2018 07:08:28
%S -1,0,-1,1,0,-2,2,6,0,-6,3,20,35,0,-24,4,45,170,225,0,-120,5,84,525,
%T 1470,1624,0,-720,6,140,1288,5880,13538,13132,0,-5040,7,216,2730,
%U 18144,67347,134568,118124,0,-40320,8,315,5220,47250,253092,807975,1447360,1172700,0,-362880,9,440,9240,108900,788865,3608220,10250790,16819000,12753576,0,-3628800
%N Triangle read by rows: T(n,k) = (n-1-k)*abs(s(n,n+1-k)), where s(n,k) are the signed Stirling numbers of the first kind and 1 <= k <= n.
%C Use the T(n,k) as coefficients to generate a polynomial of degree n-1 in d as Sum_{k=1..n} T(n,k)d^(k-1) and let g(n) be the greatest root of this polynomial. Then a polygon of n sides that form a harmonic progression in the ratio 1 : 1/(1+d) : 1/(1+2d) : ... : 1/(1+(n-1)d) can only exist if the common difference d of the denominators is limited to the range f(n) < d < g(n). The lower limit f(n) is the greatest root of another group of polynomials defined by coefficients in the triangle A199221.
%F The triangle of coefficients can be generated by expanding the equation (Sum_{k=1..n} 1/(1+(k-1)*d)) - 2 = 0 into a polynomial of degree n-1 in d.
%e Triangle starts:
%e -1;
%e 0, -1;
%e 1, 0, -2;
%e 2, 6, 0, -6;
%e 3, 20, 35, 0, -24;
%e 4, 45, 170, 225, 0, -120;
%t Flatten[Table[(n-1-k)Abs[StirlingS1[n,n+1-k]],{n,1,20},{k,1,n}]]
%o (PARI) T(n,k) = (n-1-k)*abs(stirling(n,n+1-k,1)); \\ _Michel Marcus_, Sep 30 2018
%Y Cf. A094638, A192918, A199221.
%K sign,tabl
%O 1,6
%A _Frank M Jackson_, Nov 04 2011
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