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A357805 a(n) = coefficient of x^(4*n)/(4*n)! in power series C(x) = 1 + Integral S(x)^3 * C(x)^3 dx such that C(x)^4 - S(x)^4 = 1.

%I A357805 #12 Dec 03 2022 12:01:59
%S A357805 1,6,8316,98843976,4698140798736,623259279912288096,
%T A357805 186936162949832833285056,110352751044119383032310847616,
%U A357805 116215132158682166284921510741483776,202905498509713715271588290261091671041536,554890365215965228675768455367962915432839248896
%N A357805 a(n) = coefficient of x^(4*n)/(4*n)! in power series C(x) = 1 + Integral S(x)^3 * C(x)^3 dx such that C(x)^4 - S(x)^4 = 1.
%C A357805 Equals row sums of triangles A357801 and A357802.
%H A357805 Paul D. Hanna, <a href="/A357805/b357805.txt">Table of n, a(n) for n = 0..200</a>
%F A357805 Generating function C(x) = Sum_{n>=0} a(n)*x^(4*n)/(4*n)! and related function S(x) satisfies the following formulas.
%F A357805 For brevity, some formulas here will use C = C(x) and S = S(x), where S(x) = (C(x)^4 - 1)^(1/4) is the e.g.f. of A357804.
%F A357805 (1) C(x)^4 - S(x)^4 = 1.
%F A357805 Integral formulas.
%F A357805 (2.a) S(x) = Integral C(x)^6 dx.
%F A357805 (2.b) C(x) = 1 + Integral S(x)^3 * C(x)^3 dx.
%F A357805 (2.c) S(x)^4 = Integral 4 * S(x)^3 * C(x)^6 dx.
%F A357805 (2.d) C(x)^4 = 1 + Integral 4 * S(x)^3 * C(x)^6 dx.
%F A357805 Derivatives.
%F A357805 (3.a) d/dx S(x) = C(x)^6.
%F A357805 (3.b) d/dx C(x) = S(x)^3 * C(x)^3.
%F A357805 Exponential formulas.
%F A357805 (4.a) C + S = exp( Integral (C^2 - C*S + S^2) * C^3 dx ).
%F A357805 (4.b) C - S = exp( -Integral (C^2 + C*S + S^2) * C^3 dx ).
%F A357805 (5.a) C^2 + S^2 = exp( 2 * Integral S*C^4 dx ).
%F A357805 (5.b) C^2 - S^2 = exp( -2 * Integral S*C^4 dx ).
%F A357805 Hyperbolic functions.
%F A357805 (6.a) C = sqrt(C^2 - S^2) * cosh( Integral (C^2 + S^2) * C^3 dx ).
%F A357805 (6.b) S = sqrt(C^2 - S^2) * sinh( Integral (C^2 + S^2) * C^3 dx ).
%F A357805 (7.a) C^2 = cosh( 2 * Integral S*C^4 dx ).
%F A357805 (7.b) S^2 = sinh( 2 * Integral S*C^4 dx ).
%F A357805 Explicit formulas.
%F A357805 (8.a) S(x) = Series_Reversion( Integral 1/(1 + x^4)^(3/2) dx ).
%F A357805 (8.b) C( Integral 1/(1 + x^4)^(3/2) dx ) = (1 + x^4)^(1/4).
%e A357805 E.g.f.: 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! + 110352751044119383032310847616*x^28/28! + ...
%e A357805 such that
%e A357805 C( Integral 1/(1 + x^4)^(3/2) dx ) = (1 + x^4)^(1/4)
%e A357805 also
%e A357805 C(x)^4 - S(x)^4 = 1,
%e A357805 where
%e A357805 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! + ... + A357804(n)*x^(4*n+1)/(4*n+1)! + ...
%o A357805 (PARI) /* Using Series Reversion (faster) */
%o A357805 {a(n) = my(S = serreverse( intformal( 1/(1 + x^4 +O(x^(4*n+4)))^(3/2) )) );
%o A357805 (4*n)!*polcoeff( (1 + S^4)^(1/4), 4*n)}
%o A357805 for(n=0, 10, print1( a(n), ", "))
%o A357805 (PARI) {a(n) = my(S=x, C=1); for(i=0, n,
%o A357805 S = intformal( C^6 +O(x^(4*n+4)));
%o A357805 C = 1 + intformal( S^3*C^3 ) );
%o A357805 (4*n)!*polcoeff( C, 4*n)}
%o A357805 for(n=0, 10, print1( a(n), ", "))
%Y A357805 Cf. A357804 (S(x)), A357801, A357802, A153300.
%K A357805 nonn
%O A357805 0,2
%A A357805 _Paul D. Hanna_, Oct 14 2022