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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. 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 (list; table; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A369408 A136202 A075418 * A096870 A350149 A350609
KEYWORD
sign,tabl
AUTHOR
Frank M Jackson, Nov 04 2011
STATUS
approved

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Last modified July 31 19:34 EDT 2024. Contains 374808 sequences. (Running on oeis4.)