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A199220 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. 3
-1, 0, -1, 1, 0, -2, 2, 6, 0, -6, 3, 20, 35, 0, -24, 4, 45, 170, 225, 0, -120, 5, 84, 525, 1470, 1624, 0, -720, 6, 140, 1288, 5880, 13538, 13132, 0, -5040, 7, 216, 2730, 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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,6
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 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.
LINKS
FORMULA
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.
EXAMPLE
Triangle starts:
-1;
0, -1;
1, 0, -2;
2, 6, 0, -6;
3, 20, 35, 0, -24;
4, 45, 170, 225, 0, -120;
MATHEMATICA
Flatten[Table[(n-1-k)Abs[StirlingS1[n, n+1-k]], {n, 1, 20}, {k, 1, n}]]
PROG
(PARI) T(n, k) = (n-1-k)*abs(stirling(n, n+1-k, 1)); \\ Michel Marcus, Sep 30 2018
CROSSREFS
Sequence in context: A011144 A127649 A274440 * A047916 A101207 A186435
KEYWORD
sign,tabl
AUTHOR
Frank M Jackson, Nov 04 2011
STATUS
approved

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Last modified March 2 15:00 EST 2024. Contains 370493 sequences. (Running on oeis4.)