A318110 Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.

0, 1, 1, 3, 3, 1, 26, 26, 11, 2, 367, 367, 167, 42, 5, 7142, 7142, 3352, 944, 163, 14, 176766, 176766, 84308, 25006, 4965, 638, 42, 5304356, 5304356, 2554329, 779246, 165474, 24924, 2510, 132, 186954535, 186954535, 90600599, 28120586, 6200455, 1010814, 121086, 9908, 429, 7566084686, 7566084686, 3683084984, 1156456088, 261067596, 44535120, 5829880, 574128, 39203, 1430

Offset: 0

Author

Gheorghe Coserea, Sep 05 2018

Keywords

• nonn
• tabl

Examples

			A(x,t) = (1+t)*x + (3+3*t+t^2)*x^2 + (26+26*t+11*t^2+2*t^3)*x^3 + ...
Triangle starts:
n\k [0]       [1]       [2]      [3]      [4]     [5]     [6]    [7]  [8]
[0] 0;
[1] 1,        1;
[2] 3,        3,        1;
[3] 26,       26,       11,      2;
[4] 367,      367,      167,     42,      5;
[5] 7142,     7142,     3352,    944,     163,    14;
[6] 176766,   176766,   84308,   25006,   4965,   638,    42;
[7] 5304356,  5304356,  2554329, 779246,  165474, 24924,  2510,  132;
[8] 186954535,186954535,90600599,28120586,6200455,1010814,121086,9908,429;
[9] ...
		

Links

• Gheorghe Coserea, Rows n=0..100, flattened
• Noam Zeilberger, Counting isomorphism classes of beta-normal linear lambda terms, arXiv:1509.07596 [cs.LO], 2015.

Crossrefs

Column 0 gives A262301.

Main diagonal gives A000108(n-1) for n>0.

Second diagonal gives A032443(n-1) for n>0.

Programs

Mathematica

rows = 10; Clear[A]; A[x_, t_] = (1+t)x;
Do[A[x_, t_] = Series[x t/(1-A[x, t]) + D[A[x, t], t], {x, 0, n}, {t, 0, n}] // Normal, {n, 2 rows}];
CoefficientList[#, t]& /@ CoefficientList[A[x, t], x] /. {} -> {0} // Take[#, rows]& // Flatten (* Jean-François Alcover, Oct 23 2018 *)

PARI

seq(N) = {
  my(x='x+O('x^N), t='t, F0=(1+t)*x, F1=0, n=1);
  while(n++,
    F1 = F0^2; F1 = F1 - deriv(F1,'t)/2 + deriv(F0,'t) + x*t;
    if (F1 == F0, break()); F0 = F1);
  concat([[0]], apply(Vecrev, Vec(F0)));
};
concat(seq(10))
\\ test: y=Ser(apply(p->Polrev(p,'t), seq(101)), 'x); y == x*'t/(1-y) + deriv(y,'t)

Formula

A(x,t) = Sum_{n>=0} P_n(t)*x^n, where P_n(t) = Sum_{k=0..n} T(n,k)*t^k, satisfies:

A = x*t/(1-A) + deriv(A,t), with A(0,t) = 0, deriv(A,x)(0,t) = 1+t (deriv(A,v) represents the derivative of A with respect to variable v).