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%I #26 Sep 30 2018 07:08:17
%S -1,0,1,1,4,2,2,12,18,6,3,28,83,88,24,4,55,270,575,500,120,5,96,705,
%T 2490,4324,3288,720,6,154,1582,8330,23828,35868,24696,5040,7,232,3178,
%U 23296,98707,242872,328236,209088,40320,8,333,5868,57078,334740,1212057,2658472,3298932,1972512,362880,9,460,10140,126300,977865,4873680,15637290,31292600,36207576,20531520,3628800
%N Triangle read by rows: T(n,k) = (n+1-k)*|s(n,n+1-k)| - 2*|s(n-1,n-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 f(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 higher limit g(n) is the greatest root of another group of polynomials defined by coefficients in the triangle A199220.
%F The triangle of coefficients can be generated by expanding the equation (Sum_{k=1..n} 1/(1+(k-1)d)) - 2/(1+(n-1)d) = 0 into a polynomial of degree n-1 in d.
%e Triangle starts:
%e -1;
%e 0, 1;
%e 1, 4, 2;
%e 2, 12, 18, 6;
%e 3, 28, 83, 88, 24;
%e 4, 55, 270, 575, 500, 120;
%t Flatten[Table[(n+1-k)Abs[StirlingS1[n,n+1-k]]-2Abs[StirlingS1[n-1,n-k]],{n,1,20},{k,1,n}]]
%o (PARI) T(n,k) = (n+1-k)*abs(stirling(n,n+1-k,1)) - 2*abs(stirling(n-1,n-k,1));
%o tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Sep 30 2018
%Y Cf. A094638, A192918, A199220.
%K sign,tabl
%O 1,5
%A _Frank M Jackson_, Nov 04 2011
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