A199221 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.

-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

Offset

1,5

Comments

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.

Links

Table of n, a(n) for n=1..66.

Formula

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.

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+1-k)Abs[StirlingS1[n, n+1-k]]-2Abs[StirlingS1[n-1, n-k]], {n, 1, 20}, {k, 1, n}]]

Prog

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

Crossrefs

Cf. A094638, A192918, A199220.

Sequence in context: A369408 A136202 A075418 * A096870 A350149 A350609

Adjacent sequences: A199218 A199219 A199220 * A199222 A199223 A199224

Keyword

sign,tabl

Author

Frank M Jackson, Nov 04 2011

Status

approved