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%I #15 Jul 25 2014 08:06:58
%S 1,1,1,1,2,4,12,34,120,412,1608,6244,26288,111448,499256,2265288,
%T 10701896,51339768,254175048,1278947304,6604214760,34662182904,
%U 186002333640,1014140252376,5638617162312,31837193871480,182962292354376,1067120997002680,6325487157903240,38030207563538680
%N G.f. A(x) satisfies: [x^(n+1)] A(x)^n = n*([x^(n-1)] A(x)^n) for n>=1.
%C The g.f. G(x) of the closely related sequence A245311 satisfies:
%C (1) G(x) = 1 + x*G(x) + x^2*G(x)^2 + x^3*G(x)*G'(x).
%C (2) G(x)^2 = Dx(G(x)) / (1 + x^2*Dx^2(G(x))), where Dx(F(x)) = d/dx x*F(x).
%H Paul D. Hanna, <a href="/A245310/b245310.txt">Table of n, a(n) for n = 0..500</a>
%F G.f. satisfies:
%F (1) A(x) = 1 + x + x^2 + x^3*A'(x)/(A(x) - x*A'(x)).
%F (2) A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A245311.
%F (3) A(x) = x / Series_Reversion(x*G(x)) where G(x) is the g.f. of A245311.
%e G.f.: A(x) = 1 + x + x^2 + x^3 + 2*x^4 + 4*x^5 + 12*x^6 + 34*x^7 +...
%e The table of coefficients of x^k in A(x)^n begin:
%e n=1: [1, 1, 1, 1, 2, 4, 12, 34, 120, 412, 1608, ...];
%e n=2: [1, 2, 3, 4, 7, 14, 37, 104, 344, 1172, 4412, ...];
%e n=3: [1, 3, 6, 10, 18, 36, 88, 240, 753, 2515, 9174, ...];
%e n=4: [1, 4, 10, 20, 39, 80, 188, 496, 1487, 4836, 17122, ...];
%e n=5: [1, 5, 15, 35, 75, 161, 375, 965, 2785, 8795, 30241, ...];
%e n=6: [1, 6, 21, 56, 132, 300, 708, 1800, 5046, 15484, 51738, ...];
%e n=7: [1, 7, 28, 84, 217, 525, 1274, 3242, 8918, 26684, 86772, ...];
%e n=8: [1, 8, 36, 120, 338, 872, 2196, 5656, 15423, 45248, 143576, ...];
%e n=9: [1, 9, 45, 165, 504, 1386, 3642, 9576, 26127, 75655, 235143, ...]; ...
%e from which we can illustrate [x^(n+1)] A(x)^n = n*([x^(n-1)] A(x)^n):
%e n=1: [x^2] A(x) = 1 = 1*([x^0] A(x)) = 1*1 ;
%e n=2: [x^3] A(x)^2 = 4 = 2*([x^1] A(x)^2) = 2*2 ;
%e n=3: [x^4] A(x)^3 = 18 = 3*([x^2] A(x)^3) = 3*6 ;
%e n=4: [x^5] A(x)^4 = 80 = 4*([x^3] A(x)^4) = 4*20 ;
%e n=5: [x^6] A(x)^5 = 375 = 5*([x^4] A(x)^5) = 5*75 ;
%e n=6: [x^7] A(x)^6 = 1800 = 6*([x^5] A(x)^6) = 6*300 ;
%e n=7: [x^8] A(x)^7 = 8918 = 7*([x^6] A(x)^7) = 7*1274 ;
%e n=8: [x^9] A(x)^8 = 45248 = 8*([x^7] A(x)^8) = 8*5656 ;
%e n=9: [x^10] A(x)^9 = 235143 = 9*([x^8] A(x)^9) = 9*26127 ; ...
%e describing terms that lie along diagonals in the above table.
%e From the main diagonal in the above table, we may derive A245311:
%e [1/1, 2/2, 6/3, 20/4, 75/5, 300/6, 1274/7, 5656/8, 26127/9, ...]
%e = [1, 1, 2, 5, 15, 50, 182, 707, 2903, 12479, ...].
%o (PARI) /* From A(x) = 1+x+x^2 + x^3*A'(x)/(A(x) - x*A'(x)): */
%o {a(n)=local(A=1+x); for(i=1, n, A = 1+x+x^2 + x^3*A'/(A-x*A' +x*O(x^n))); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) /* From [x^(n-1)] A(x)^n = n*([x^(n+1)] A(x)^n): */
%o {a(n)=local(A=1+x+x^2);for(m=2,n,A=A-x^(m+1)*(polcoeff(A^m+O(x^(m+2)),m+1)/m - polcoeff(A^m+O(x^m),m-1))+O(x^(n+2)));polcoeff(A,n)}
%o for(n=0,30,print1(a(n),", "))
%o (PARI) /* From the g.f. of A245311: */
%o {a(n)=local(G=1+x); for(i=1, n, G = (1 + x^2*G^2 + x^3*G*G')/(1-x +x*O(x^n)));
%o polcoeff(x/serreverse(x*G +x^2*O(x^n)), n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A245311, A245312.
%K nonn
%O 0,5
%A _Paul D. Hanna_, Jul 17 2014