%I #12 Dec 03 2022 12:03:38
%S 1,-1,1,107,913,-131449,-46887791,4109309363,406392278497,
%T 295047521858639,5615320767861601,-121434328185686247493,
%U 13788915057049470673393,30743837939769538654859351,-10050889695209166245600514191,-2332393553453526728340631941757
%N a(n) = coefficient of x^(2*n)/(2*n)! in the expansion of the even function C(x) = sqrt(1 - S(x)^2) where S(x) is defined by A357550.
%H Paul D. Hanna, <a href="/A357551/b357551.txt">Table of n, a(n) for n = 0..300</a>
%F G.f. C(x) = Sum_{n>=0} a(n) * x^(2*n)/(2*n)! and related odd function S(x) satisfies the following relations.
%F (1) S(x) = Integral Product_{n>=1} (1 - S(x)^(2*n))^((2*n-1)/(2*n)) dx.
%F (2) C(x) = 1 - Integral S(x) * Product_{n>=2} (1 - S(x)^(2*n))^((2*n-1)/(2*n)).
%F (3) C(x) = sqrt(1 - S(x)^2).
%F (4) C'(x)*C(x) = -S'(x)*S(x).
%F (5) C(x) + i*S(x) = exp( i * Integral S'(x) / C(x) dx ), where i^2 = -1.
%F (5.a) C(x) = cos( Integral S'(x) / C(x) dx ).
%F (5.b) S(x) = sin( Integral S'(x) / C(x) dx ).
%F (6) C(x)^2 - S(x)^2 = cos( 2 * Integral S'(x) / C(x) dx ).
%F (6.a) 2*C(x)^2 = 1 + cos( 2 * Integral S'(x) / C(x) dx ).
%F (6.b) 2*S(x)^2 = 1 - cos( 2 * Integral S'(x) / C(x) dx ).
%F (7) S(x) = Series_Reversion( Integral Product_{n>=1} (1 - x^(2*n))^(-(2*n-1)/(2*n)) dx ).
%e G.f.: C(x) = 1 - x^2/2! + x^4/4! + 107*x^6/6! + 913*x^8/8! - 131449*x^10/10! - 46887791*x^12/12! + 4109309363*x^14/14! + 406392278497*x^16/16! + ...
%e where C(x) = sqrt(1 - S^2) and
%e S(x) = x - x^3/3! - 17*x^5/5! + 137*x^7/7! + 13009*x^9/9! + 3098111*x^11/11! - 499973633*x^13/13! + 13063051433*x^15/15! + ... + A357550(n)*x^(2*n-1)/(2*n-1)! + ...
%o (PARI) {a(n) = my(S); S = serreverse( intformal( prod(k=1, n+1, (1 - x^(2*k) + x*O(x^(2*n)) )^(-(2*k-1)/(2*k)) ) ) ); (2*n)! * polcoeff( sqrt(1 - S^2), 2*n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) {a(n) = my(S=x); for(i=1, n, S = intformal( prod(k=1, n+1, (1 - S^(2*k) + x*O(x^(2*n)))^((2*k-1)/(2*k)) ) ) ); (2*n)! * polcoeff( sqrt(1 - S^2), 2*n)}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A357550, A357231.
%K sign
%O 0,4
%A _Paul D. Hanna_, Oct 04 2022