%I #14 Feb 23 2018 03:41:23
%S 0,0,1,2,1,1,-1,1,1,-139,1,-571,-281,163879,-5221,5246819,5459,
%T -534703531,91207079,-4483131259,-2650986803,432261921612371,
%U -6171801683,6232523202521089,4283933145517,-25834629665134204969,11963983648109,-1579029138854919086429,-208697624924077,746590869962651602203151,-29320119130515566117
%N Numerators of coefficients in S(x) where: C(x)^(1/2) - S(x)^(1/2) = 1 such that C'(x)*S(x)^(1/2) = S'(x)*C(x)^(1/2) = 2*x*C(x).
%F The functions C = C(x) and S = S(x) satisfy:
%F (1) sqrt(C) - sqrt(S) = 1.
%F (2a) C'*sqrt(S) = S'*sqrt(C) = 2*x*C.
%F (2b) C' = 2*x*C/sqrt(S).
%F (2c) S' = 2*x*sqrt(C).
%F (3a) C = 1 + Integral 2*x*C/sqrt(S) dx.
%F (3b) S = Integral 2*x*sqrt(C) dx.
%F (4a) sqrt(C) = exp( Integral x/(sqrt(C) - 1) dx ).
%F (4b) sqrt(S) = exp( Integral x/sqrt(S) dx ) - 1.
%F (5a) C - S = exp( Integral 2*x*C/(C*sqrt(S) + S*sqrt(C)) dx ).
%F (5b) C - S = exp( Integral C'*S'/(C*S' + S*C') dx).
%F (6a) sqrt(C) = exp( sqrt(C) - 1 - x^2/2 ).
%F (6b) sqrt(C) = 1 + x^2/2 + Integral x/(sqrt(C) - 1) dx.
%e G.f.: S(x) = x^2 + 2/3*x^3 + 1/6*x^4 + 1/90*x^5 - 1/810*x^6 + 1/15120*x^7 + 1/68040*x^8 - 139/24494400*x^9 + 1/1020600*x^10 - 571/12933043200*x^11 + ...
%e Related power series begin:
%e C(x) = 1 + 2*x + 5/3*x^2 + 13/18*x^3 + 43/270*x^4 + 5/432*x^5 - 19/17010*x^6 + 41/2721600*x^7 + 1/40824*x^8 - 7243/1175731200*x^9 + 923/1515591000*x^10 + ...
%e sqrt(C) = 1 + x + 1/3*x^2 + 1/36*x^3 - 1/270*x^4 + 1/4320*x^5 + 1/17010*x^6 - 139/5443200*x^7 + 1/204120*x^8 - 571/2351462400*x^9 - 281/1515591000*x^10 + ... + A005447(n)/A005446(n)*x^n + ...
%t terms = 30; c[x_] = Assuming[x > 0, ProductLog[-1, -Exp[-1 - x^2/2]]^2 + O[x]^terms]; Integrate[2*x*Sqrt[c[x]] + O[x]^terms, x] // CoefficientList[#, x] & // Numerator (* _Jean-François Alcover_, Feb 22 2018 *)
%o (PARI) {a(n) = my(C=1, S=x^2); for(i=0,n, C = (1 + sqrt(S +O(x^(n+2))))^2; S = intformal( 2*x*sqrt(C) ) ); numerator(polcoeff(S,n))}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A299433 (denominators in S), A299430/A299431 (C), A005447/A005446 (sqrt(C).
%K sign,frac
%O 0,4
%A _Paul D. Hanna_, Feb 09 2018