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%I #9 Mar 30 2012 18:37:33
%S 1,1,3,19,183,2311,35523,637015,12978591,295150015,7397981283,
%T 202454752039,6004327533471,191822535178639,6567996718854483,
%U 239978667532009159,9321006255172812543,383560500541816176319,16671266628681878467011,763268479436848970368615
%N G.f. A(x) satisfies: a(n) = [x^(n-1)] A( x/(1-x) )^n / (1-x) for n>0 with a(0)=1 where A(x) = Sum_{n>=0} a(n)*x^n.
%F a(n) equals the coefficient of x^(n-1) in the binomial transform of A(x)^n for n>0 with a(0)=1 where A(x) is the g.f. of this sequence.
%e G.f.: A(x) = 1 + x + 3*x^2 + 19*x^3 + 183*x^4 + 2311*x^5 + 35523*x^6 +...
%e where the coefficients in the initial powers A(x)^n begin:
%e n=1: [1, 1, 3, 19, 183, 2311, 35523, 637015, ...];
%e n=2: [1, 2, 7, 44, 413, 5102, 77127, 1365896, ...];
%e n=3: [1, 3, 12, 76, 699, 8457, 125730, 2198466, ...];
%e n=4: [1, 4, 18, 116, 1051, 12472, 182380, 3148072, ...];
%e n=5: [1, 5, 25, 165, 1480, 17256, 248270, 4229770, ...];
%e n=6: [1, 6, 33, 224, 1998, 22932, 324754, 5460528, ...];
%e n=7: [1, 7, 42, 294, 2618, 29638, 413364, 6859448, ...];
%e n=8: [1, 8, 52, 376, 3354, 37528, 515828, 8448008, ...]; ...
%e The coefficients in the binomial transform of the powers of A(x) begin:
%e n=1: [(1), 2, 6, 32, 282, 3452, 52566, 941348, ...];
%e n=2: [1,(3), 12, 72, 640, 7688, 114932, 2029084, ...];
%e n=3: [1, 4,(19), 122, 1088, 12848, 188676, 3283572, ...];
%e n=4: [1, 5, 27,(183), 1640, 19088, 275592, 4727896, ...];
%e n=5: [1, 6, 36, 256,(2311), 26582, 377712, 6388172, ...];
%e n=6: [1, 7, 46, 342, 3117,(35523), 497328, 8293884, ...];
%e n=7: [1, 8, 57, 442, 4075, 46124,(637015), 10478246, ...];
%e n=8: [1, 9, 69, 557, 5203, 58619, 799655, (12978591), ...]; ...
%e where the coefficients in parenthesis form the initial terms of this sequence.
%o (PARI) {a(n)=local(A=1+x+sum(k=2,n-1,a(k)*x^k)+x*O(x^n));if(n==0,1,sum(k=0,n-1,binomial(n-1,k)*polcoeff(A^n,k)))}
%o (PARI) {a(n)=local(A=1+x+sum(k=2,n-1,a(k)*x^k)+x*O(x^n));if(n==0,1,polcoeff(subst(A^n,x,x/(1-x+x*O(x^n)))/(1-x),n-1))}
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 29 2011
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