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A190392 E.g.f. A(x) satisfies A'(x) = sin(A(x)) + cos(A(x)).

%I A190392 #57 Jan 15 2018 03:10:30
%S A190392 1,1,0,-4,-12,4,240,1184,-1008,-59504,-401280,643136,38584128,
%T A190392 323581504,-848090880,-51666451456,-509739310848,2004840714496,
%U A190392 123888658698240,1386061762251776,-7721141999864832,-483475390212586496,-5974101514137292800,45231727252157947904
%N A190392 E.g.f. A(x) satisfies A'(x) = sin(A(x)) + cos(A(x)).
%C A190392 Let f(x) be a smooth function. The autonomous differential equation A'(x) = f(A(x)), with initial condition A(0) = 0, is separable and the solution is given by A(x) = inverse function of Integral_{t = 0..x} 1/f(t) dt. The inversion of the integral Integral_{t = 0..x} 1/f(t) dt is most conveniently found by applying [Dominici, Theorem 4.1]. The result is A(x) = Sum_{n>=1} D^(n-1)[f](0)*x^n/n!, where the nested derivative D^n[f](x)is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = (d/dx)(f(x)*D^n[f](x)) for n >= 0. See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x). In the present case we take f(x) = sin(x)+cos(x). - _Peter Bala_, Aug 27 2011
%H A190392 Alois P. Heinz, <a href="/A190392/b190392.txt">Table of n, a(n) for n = 1..200</a>
%H A190392 D. Dominici, <a href="http://arxiv.org/abs/math/0501052">Nested derivatives: A simple method for computing series expansions of inverse functions.</a> arXiv:math/0501052v2 [math.CA], 2005.
%F A190392 E.g.f.: A(x) = inverse of Integral_{t = 0..x} 1/(sin(t)+cos(t)) dt = series reversion (x - x^2/2! + 3*x^3/3! - 11*x^4/4! + 57*x^5/5! - ...) = x + x^2/2! - 4*x^4/4! - 12*x^5/5! + .... a(n) = D^(n-1)[sin(x)+cos(x)](0), where the nested derivative operator D^n is defined above. Compare with A012244. -_Peter Bala_, Aug 27 2011
%F A190392 E.g.f.: A(x) = 2*arctan((sqrt(2)-1)*exp(sqrt(2)*x))-Pi/4. Compare with A028296. - _Peter Bala_, Sep 02 2011
%F A190392 G.f.: 1/G(0) where G(k) = 1 - 2*x*(k+1)/(1 + 1/(1 - 2*x*(k+1)/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Jan 10 2013.
%F A190392 G.f.: -(1/x)/Q(0), where Q(k)= 2*k+1 - 1/x + (k+1)*(k+1)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Apr 15 2013
%F A190392 G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 + (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Oct 17 2013
%F A190392 E.g.f. if offset 0: 2^(1/2)/(2^(1/2)*cosh(x*2^(1/2))-sinh(x*2^(1/2))). - _Sergei N. Gladkovskii_, Nov 10 2013
%F A190392 a(n) ~ -(n-1)! * 2^(3*n/2+1) * sin(n*arctan(Pi/log(3 - 2*sqrt(2)))) / (Pi^2 + log(3 - 2*sqrt(2))^2)^(n/2). - _Vaclav Kotesovec_, Jan 07 2014
%p A190392 A := x ; for i from 1 to 35 do sin(A)+cos(A) ; convert(taylor(%,x=0,25),polynom) ; A := int(%,x) ; print(A) ; end do:
%p A190392 for i from 1 to 25 do printf("%d,", coeftayl(A,x=0,i)*i!) ; end do: # _R. J. Mathar_, Jun 03 2011
%t A190392 terms = 25; A[_] = 0; Do[A[x_] = Integrate[Sin[A[x]] + Cos[A[x]], x] + O[x]^terms, terms]; CoefficientList[A[x], x]*Range[0, terms-1]! (* _Jean-François Alcover_, Feb 21 2013, updated Jan 15 2018 *)
%o A190392 (Maxima)
%o A190392 g(n):=(-1)^floor(n/2)*1/n!;
%o A190392 a(n):=T190015_Solve(n,g);
%Y A190392 Cf. A001586, A012244, A028296.
%K A190392 sign
%O A190392 1,4
%A A190392 _Vladimir Kruchinin_, May 09 2011