

A199221


Triangle read by rows: T(n,k) = (n+1k)*s(n,n+1k)  2*s(n1,nk), where s(n,k) are the signed Stirling numbers of the first kind and 1 <= k <= n.


3



1, 0, 1, 1, 4, 2, 2, 12, 18, 6, 3, 28, 83, 88, 24, 4, 55, 270, 575, 500, 120, 5, 96, 705, 2490, 4324, 3288, 720, 6, 154, 1582, 8330, 23828, 35868, 24696, 5040, 7, 232, 3178, 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
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OFFSET

1,5


COMMENTS

Use the T(n,k) as coefficients to generate a polynomial of degree n1 in d as Sum_{k=1..n} T(n,k)d^(k1) 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+(n1)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.


LINKS



FORMULA

The triangle of coefficients can be generated by expanding the equation (Sum_{k=1..n} 1/(1+(k1)d))  2/(1+(n1)d) = 0 into a polynomial of degree n1 in d.


EXAMPLE

Triangle starts:
1;
0, 1;
1, 4, 2;
2, 12, 18, 6;
3, 28, 83, 88, 24;
4, 55, 270, 575, 500, 120;


MATHEMATICA

Flatten[Table[(n+1k)Abs[StirlingS1[n, n+1k]]2Abs[StirlingS1[n1, nk]], {n, 1, 20}, {k, 1, n}]]


PROG

(PARI) T(n, k) = (n+1k)*abs(stirling(n, n+1k, 1))  2*abs(stirling(n1, nk, 1));
tabl(nn) = for (n=1, nn, for (k=1, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Sep 30 2018


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



