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Triangle of coefficients, read by rows of (2n+1) terms, where the n-th row forms a polynomial in x, P(n,x), of degree 2n and satisfies: P(n,x) = [Sum_{k=1..n} 1/(k + x + x^2)]*[Product_{k=1..n} (k + x + x^2)].
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%I #24 Aug 06 2017 22:20:44

%S 1,3,2,2,11,12,15,6,3,50,70,100,64,42,12,4,274,450,705,570,440,200,90,

%T 20,5,1764,3248,5453,5110,4410,2526,1360,480,165,30,6,13068,26264,

%U 46571,48454,45437,30128,18347,8162,3395,980,273,42,7

%N Triangle of coefficients, read by rows of (2n+1) terms, where the n-th row forms a polynomial in x, P(n,x), of degree 2n and satisfies: P(n,x) = [Sum_{k=1..n} 1/(k + x + x^2)]*[Product_{k=1..n} (k + x + x^2)].

%C These polynomials have zeros at complex z_k such that real(z_k) = -1/2 for all 0 < k < (2n-1), n > 1. A pair of zeros that are complex rationals occur at n = 2k(k+1) and have the values z = -1/2 +- (2k+1)/2*i for k > 0. P(n,0) = Stirling numbers of the first kind and the product of all the zeros of P(n,x) equals P(n,0)/n!.

%H Vincenzo Librandi, <a href="/A074248/b074248.txt">Rows n = 1..32, flattened</a>

%e P(1,x) = 1,

%e P(2,x) = 3 + 2x + 2x^2,

%e P(3,x) = 11 + 12x + 15x^2 + 6x^3 + 3x^4,

%e P(4,x) = 50 + 70x + 100x^2 + 64x^3 + 42x^4 + 12x^5 + 4x^6,

%e P(5,x) = 274 + 450x + 705x^2 + 570x^3 + 440x^4 + 200x^5 + 90x^6 + 20x^7 + 5x^8,

%e P(6,x) = 1764 + 3248x + 5453x^2 + 5110x^3 + 4410x^4 + 2526x^5 + 1360x^6 + 480x^7 + 165x^8 + 30x^9 + 6x^10,

%e P(7,x) = 13068 + 26264x + 46571x^2 + 48454x^3 + 45437x^4 + 30128x^5 + 18347x^6 + 8162x^7 + 3395x^8 + 980x^9 + 273x^10 + 42x^11 + 7x^12.

%t p[n_, x_] := Sum[1/(k + x + x^2), {k, 1, n}]*Product[k + x + x^2, {k, 1, n}]; row[n_] := CoefficientList[ Series[p[n, x], {x, 0, 2*n-2}], x]; Table[row[n], {n, 1, 7}] // Flatten (* _Jean-François Alcover_, Aug 16 2013 *)

%K easy,nice,nonn,tabf

%O 1,2

%A _Paul D. Hanna_, Sep 20 2002

%E Keyword tabf by _Michel Marcus_, Aug 06 2017