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A357804 a(n) = coefficient of x^(4*n+1)/(4*n+1)! in power series S(x) = Series_Reversion( Integral 1/(1 + x^4)^(3/2) dx ).

%I A357804 #17 Apr 09 2025 06:34:42
%S A357804 1,36,87696,1483707456,91329084354816,14862901723860427776,
%T A357804 5279211177231308343054336,3600188413031639396548043882496,
%U A357804 4300014195136238449156877005063520256,8394333803654997846112872487491938363375616,25378508500092778024069322428694679252236239896576
%N A357804 a(n) = coefficient of x^(4*n+1)/(4*n+1)! in power series S(x) = Series_Reversion( Integral 1/(1 + x^4)^(3/2) dx ).
%C A357804 Equals row sums of triangle A357800.
%H A357804 Paul D. Hanna, <a href="/A357804/b357804.txt">Table of n, a(n) for n = 0..200</a>
%F A357804 Generating function S(x) = Sum_{n>=0} a(n)*x^(4*n+1)/(4*n+1)! and related function C(x) satisfies the following formulas.
%F A357804 For brevity, some formulas here will use S = S(x) and C = C(x), where C(x) = (1 + S(x)^4)^(1/4) is the e.g.f. of A357805.
%F A357804 (1) C(x)^4 - S(x)^4 = 1.
%F A357804 Integral formulas.
%F A357804 (2.a) S(x) = Integral C(x)^6 dx.
%F A357804 (2.b) C(x) = 1 + Integral S(x)^3 * C(x)^3 dx.
%F A357804 (2.c) S(x)^4 = Integral 4 * S(x)^3 * C(x)^6 dx.
%F A357804 (2.d) C(x)^4 = 1 + Integral 4 * S(x)^3 * C(x)^6 dx.
%F A357804 Derivatives.
%F A357804 (3.a) d/dx S(x) = C(x)^6.
%F A357804 (3.b) d/dx C(x) = S(x)^3 * C(x)^3.
%F A357804 Exponential formulas.
%F A357804 (4.a) C + S = exp( Integral (C^2 - C*S + S^2) * C^3 dx ).
%F A357804 (4.b) C - S = exp( -Integral (C^2 + C*S + S^2) * C^3 dx ).
%F A357804 (5.a) C^2 + S^2 = exp( 2 * Integral S*C^4 dx ).
%F A357804 (5.b) C^2 - S^2 = exp( -2 * Integral S*C^4 dx ).
%F A357804 Hyperbolic functions.
%F A357804 (6.a) C = sqrt(C^2 - S^2) * cosh( Integral (C^2 + S^2) * C^3 dx ).
%F A357804 (6.b) S = sqrt(C^2 - S^2) * sinh( Integral (C^2 + S^2) * C^3 dx ).
%F A357804 (7.a) C^2 = cosh( 2 * Integral S*C^4 dx ).
%F A357804 (7.b) S^2 = sinh( 2 * Integral S*C^4 dx ).
%F A357804 Explicit formulas.
%F A357804 (8.a) S(x) = Series_Reversion( Integral 1/(1 + x^4)^(3/2) dx ).
%F A357804 (8.b) C( Integral 1/(1 + x^4)^(3/2) dx ) = (1 + x^4)^(1/4).
%e A357804 E.g.f.: S(x) = x + 36*x^5/5! + 87696*x^9/9! + 1483707456*x^13/13! + 91329084354816*x^17/17! + 14862901723860427776*x^21/21! + 5279211177231308343054336*x^25/25! + ...
%e A357804 such that
%e A357804 S( Integral 1/(1 + x^4)^(3/2) dx ) = x
%e A357804 also
%e A357804 C(x)^4 - S(x)^4 = 1,
%e A357804 where
%e A357804 C(x) = 1 + 6*x^4/4! + 8316*x^8/8! + 98843976*x^12/12! + 4698140798736*x^16/16! + 623259279912288096*x^20/20! + 186936162949832833285056*x^24/24! + ... + A357805(n)*x^(4*n)/(4*n)! + ...
%t A357804 nmax = 20; Select[CoefficientList[InverseSeries[Series[x*(1/Sqrt[1 + x^4] + Hypergeometric2F1[1/4, 1/2, 5/4, -x^4])/2, {x, 0, 4*nmax + 4}], x], x], #1 != 0 &] * Table[(4*k+1)!, {k, 0, nmax}] (* _Vaclav Kotesovec_, Apr 09 2025 *)
%o A357804 (PARI) /* Using Series Reversion (faster) */
%o A357804 {a(n) = my(S = serreverse( intformal( 1/(1 + x^4 +O(x^(4*n+4)))^(3/2) )) );
%o A357804 (4*n+1)!*polcoeff( S, 4*n+1)}
%o A357804 for(n=0, 10, print1( a(n), ", "))
%o A357804 (PARI) {a(n) = my(S=x, C=1); for(i=0, n,
%o A357804 S = intformal( C^6 +O(x^(4*n+4)));
%o A357804 C = 1 + intformal( S^3*C^3 ) );
%o A357804 (4*n)!*polcoeff( C, 4*n)}
%o A357804 for(n=0, 10, print1( a(n), ", "))
%Y A357804 Cf. A357805 (C(x)), A357800, A153301.
%K A357804 nonn
%O A357804 0,2
%A A357804 _Paul D. Hanna_, Oct 14 2022