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Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.
2

%I #23 Sep 28 2019 12:30:02

%S 1,2,2,7,11,6,30,65,59,22,143,397,492,318,90,728,2471,3857,3430,1728,

%T 394,3876,15572,29255,32728,22886,9461,1806,21318,99009,217323,291456,

%U 257001,148626,52133,8558,120175,633765,1591231,2481597,2622445,1918665,947740,288812,41586,690690,4078360,11527318,20467755,25114375,22043890,13821764,5964728,1607198,206098

%N Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.

%H Gheorghe Coserea, <a href="/A278457/b278457.txt">Rows n = 1..101, flattened</a>

%H F. Chapoton, F. Hivert, J.-C. Novelli, <a href="http://arxiv.org/abs/1307.0092">A set-operad of formal fractions and dendriform-like sub-operads</a>, arXiv preprint arXiv:1307.0092 [math.CO], 2013.

%F y(x) = Sum {n>=1} P_n(t)*x^n satisfies x = y*((t-1)*y^3 + (t^2-2*t-1)*y^2 - (2*t-1)*y + 1)/((1-t)*y^3 + (3-t)*y^2 + 3*y + 1), with y(0)=0, y'(0)=1, where P_n(t) is the degree n-1 polynomial associated with row n of the triangle in order of decreasing powers of t.

%F P_n(0) = A006318(n-1), P_n(1) = A156017(n-1), P_n(2) = A231690(n).

%e A(x;t) = x + (2*t+2)*x^2 + (7*t^2+11*t+6)*x^3 + (30*t^3+65*t^2+59*t+22)*x^4 + ...

%e Triangle starts:

%e n\k [1] [2] [3] [4] [5] [6] [7] [8]

%e [1] 1;

%e [2] 2, 2;

%e [3] 7, 11, 6;

%e [4] 30, 65, 59, 22;

%e [5] 143, 397, 492, 318, 90;

%e [6] 728, 2471, 3857, 3430, 1728, 394;

%e [7] 3876, 15572, 29255, 32728, 22886, 9461, 1806;

%e [8] 21318, 99009, 217323, 291456, 257001, 148626, 52133, 8558;

%e [9] ...

%t Reverse[CoefficientList[#, t]]& /@ CoefficientList[InverseSeries[x ((t - 1) x^3 + (t^2 - 2t - 1) x^2 - (2t - 1) x + 1)/((1 - t) x^3 + (3 - t) x^2 + 3x + 1) + O[x]^11], x] // Flatten (* _Jean-François Alcover_, Sep 28 2019 *)

%o (PARI)

%o N=11; x ='x + O('x^N);

%o concat(apply(p->Vec(p), Vec(serreverse(Ser(x*((t-1)*x^3 + (t^2-2*t-1)*x^2 - (2*t-1)*x+1)/((1-t)*x^3 + (3-t)*x^2 + 3*x + 1), 'x)))))

%Y Column k=1 give A006013.

%K nonn,tabl

%O 1,2

%A _Gheorghe Coserea_, Jan 15 2017