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%I #11 Jul 04 2015 23:04:56
%S 1,1,3,19,201,3097,63963,1677883,53862225,2059533745,91909156083,
%T 4711414480867,273922577628057,17876478783834313,1298278981158825291,
%U 104165674946626478347,9175884838706696138145,882669439812976183138657,92284482487864563215652579,10443186091555501868233274803
%N E.g.f. A(x) satisfies: A( Integral 1/A(x)^2 dx ) = exp(x).
%F E.g.f. satisfies: A(x) = exp( Series_Reversion( Integral 1/A(x)^2 dx ) ).
%e E.g.f.: A(x) = 1 + x + 3*x^2/2! + 19*x^3/3! + 201*x^4/4! + 3097*x^5/5! + 63963*x^6/6! +...
%e Related expansions.
%e A(x)^2 = 1 + 2*x + 8*x^2/2! + 56*x^3/3! + 608*x^4/4! + 9344*x^5/5! + 190400*x^6/6! +...+ A233335(n)*(2*x)^n/n! +...
%e Integral 1/A(x)^2 dx = x - 2*x^2/2! - 8*x^4/4! - 96*x^5/5! - 1664*x^6/6! +...
%e The series reversion of the Integral 1/A(x)^2 dx equals log(A(x)) and begins:
%e log(A(x)) = x + 2*x^2/2! + 12*x^3/3! + 128*x^4/4! + 2016*x^5/5! + 42656*x^6/6! + 1145280*x^7/7! + 37563008*x^8/8! +...+ A259267(n)*x^n/n! +...
%o (PARI) {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=exp(serreverse(intformal(1/A^2+x*O(x^n)))));n!*polcoeff(A,n)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A233335, A259267.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 07 2013
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