

A199220


Triangle read by rows: T(n,k) = (n1k)*abs(s(n,n+1k)), 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
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OFFSET

1,6


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 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+(n1)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

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+(k1)*d))  2 = 0 into a polynomial of degree n1 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[(n1k)Abs[StirlingS1[n, n+1k]], {n, 1, 20}, {k, 1, n}]]


PROG

(PARI) T(n, k) = (n1k)*abs(stirling(n, n+1k, 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



