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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 Cf. A094638, A192918, A199221. Sequence in context: A011144 A127649 A274440 * A047916 A101207 A186435 Adjacent sequences:  A199217 A199218 A199219 * A199221 A199222 A199223 KEYWORD sign,tabl AUTHOR Frank M Jackson, Nov 04 2011 STATUS approved

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Last modified September 19 17:19 EDT 2021. Contains 347564 sequences. (Running on oeis4.)