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%I #8 Mar 01 2017 11:02:13
%S 1,1,17,865,88865,15335425,3993275825,1462392957025,716611617346625,
%T 452780458211706625,358439197464543820625,347486061804813737430625,
%U 404905203733448633060470625,558379985997479451541000890625,899457522079287575519574474640625,1673555570600439849976672764510390625,3562028724551236205811767300836022890625
%N E.g.f. C(x) satisfies: C(x) = cosh( Integral C(x)^4 dx ).
%F E.g.f. C(x) = ( d/dx Series_Reversion( x - x^3/3 ) )^(1/2).
%F E.g.f. C(x) = ( d/dx Series_Reversion( sin(x) - sin(x)^3/3 ) )^(1/3).
%F E.g.f. C(x) = ( d/dx Series_Reversion( sinh(x)*(2 + cosh(2*x))/(3*cosh(x)^3) ) )^(1/4).
%F E.g.f. C(x) = ( d/dx Series_Reversion( x*sqrt(1+x^2)*(3 + 2*x^2)/(3*(1 + x^2)^2) ) )^(1/5).
%F E.g.f. C(x) = d/dx Series_Reversion( Integral 1/G(x) dx ) where G(x) = e.g.f. of A281181.
%F E.g.f. C(x) = ( d/dx Series_Reversion( Integral (1 - x^2) dx ) )^(1/2).
%F E.g.f. C(x) = ( d/dx Series_Reversion( Integral cos(x)^3 dx ) )^(1/3).
%F E.g.f. C(x) = ( d/dx Series_Reversion( Integral 1/cosh(x)^4 dx ) )^(1/4).
%F E.g.f. C(x) = ( d/dx Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ) )^(1/5).
%F E.g.f. C(x) = ( d/dx Series_Reversion( Integral G(i*x)^6 dx ) )^(1/6) where G(x) = e.g.f. of A281181.
%F E.g.f. C(x) and related series S(x) (e.g.f. of A281427) satisfy:
%F (1.a) C(x)^2 - S(x)^2 = 1.
%F (1.b) C(x)^2 + S(x)^2 = 1 + Integral 4*C(x)^5*S(x) dx.
%F Integrals.
%F (2.a) S(x) = Integral C(x)^5 dx.
%F (2.b) C(x) = 1 + Integral C(x)^4*S(x) dx.
%F Exponential.
%F (3.a) C(x) + S(x) = exp( Integral C(x)^4 dx ).
%F (3.b) C(x) = cosh( Integral C(x)^4 dx ).
%F (3.c) S(x) = sinh( Integral C(x)^4 dx ).
%F Derivatives.
%F (4.a) S'(x) = C(x)^5.
%F (4.b) C'(x) = C(x)^4*S(x).
%F (4.c) (C'(x) + S'(x))/(C(x) + S(x)) = C(x)^4.
%F (4.d) (C(x)^2 + S(x)^2)' = 4*C(x)^5*S(x).
%F Explicit Solutions.
%F (5.a) S(x) = Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ).
%F (5.b) C(x)^1 = d/dx Series_Reversion( Integral 1/G(x) dx ) where G(x) = e.g.f. of A281181.
%F (5.c) C(x)^2 = d/dx Series_Reversion( Integral (1 - x^2) dx ).
%F (5.d) C(x)^3 = d/dx Series_Reversion( Integral cos(x)^3 dx ).
%F (5.e) C(x)^4 = d/dx Series_Reversion( Integral 1/cosh(x)^4 dx ).
%F (5.f) C(x)^5 = d/dx Series_Reversion( Integral 1/(1 + x^2)^(5/2) dx ).
%F (5.g) C(x)^6 = d/dx Series_Reversion( Integral G(i*x)^6 dx ) )^(1/6) where G(x) = e.g.f. of A281181.
%F (5.h) C(x)^2 = d/dx Series_Reversion( x - x^3/3 ).
%F (5.j) C(x)^3 = d/dx Series_Reversion( sin(x) - sin(x)^3/3 ).
%F (5.j) C(x)^4 = d/dx Series_Reversion( sinh(x)*(2 + cosh(2*x))/(3*cosh(x)^3) ).
%F (5.k) C(x)^5 = d/dx Series_Reversion( x*sqrt(1+x^2)*(3 + 2*x^2)/(3*(1 + x^2)^2) ).
%e E.g.f.: C(x) = 1 + x^2/2! + 17*x^4/4! + 865*x^6/6! + 88865*x^8/8! + 15335425*x^10/10! + 3993275825*x^12/12! + 1462392957025*x^14/14! + 716611617346625*x^16/16! + 452780458211706625*x^18/18! + 358439197464543820625*x^20/20! +...
%e such that
%e (1) C(x) = cosh( Integral C(x)^4 dx ),
%e (2) C(x)^2 - S(x)^2 = 1, and
%e (3) C(x) = 1 + Integral C(x)^4*S(x) dx,
%e where S(x) is described by A281427 and begins:
%e S(x) = x + 5*x^3/3! + 145*x^5/5! + 10325*x^7/7! + 1357825*x^9/9! + 284963525*x^11/11! + 87274812625*x^13/13! + 36716097543125*x^15/15! + 20309401097610625*x^17/17! + 14290053364475013125*x^19/19! +...
%t a[n_] := Module[{S = x, C = 1, C5, SC4}, For[i = 0, i <= n, i++, C5 = C^5 + x*O[x]^(2n) // Normal; S = Integrate[C5 , x]; SC4 = S*C^4+O[x]^(2n) // Normal; C = 1 + Integrate[SC4, x]]; (2n)!*Coefficient[C, x, 2n]]; Array[a, 17, 0] (* _Jean-François Alcover_, Mar 01 2017, translated from Pari *)
%o (PARI) {a(n) = my(S=x, C=1); for(i=0, n, S = intformal( C^5 +x*O(x^(2*n))); C = 1 + intformal( S*C^4 ) ); (2*n)!*polcoeff(C, 2*n)}
%o for(n=0, 30, print1(a(n), ", "))
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 21 2017
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