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%I #12 Jul 19 2024 05:59:57
%S 1,-1,0,-1,2,-1,0,0,0,-1,4,-6,6,-11,20,-21,16,-17,20,-15,6,-1,0,0,0,0,
%T 0,-1,10,-45,124,-254,472,-876,1512,-2289,3184,-4369,5880,-7491,9240,
%U -11697,15640,-22003,31916,-46450,67104,-96081,133976,-178209,226608
%N G.f. A(x) satisfies: R(x^2*A(x)) = x^3 - x^4, where R(A(x)) = x.
%H Paul D. Hanna, <a href="/A350433/b350433.txt">Table of n, a(n) for n = 1..2000</a>
%F G.f.: A(x) = Product_{n>=0} F(n), where F(0) = x, F(1) = 1-x, and F(n+1) = 1 - (1 - F(n))^3 * F(n) for n > 0.
%e G.f.: A(x) = x - x^2 - x^4 + 2*x^5 - x^6 - x^10 + 4*x^11 - 6*x^12 + 6*x^13 - 11*x^14 + 20*x^15 - 21*x^16 + 16*x^17 + ...
%e Let R(x) be the series reversion of A(x),
%e R(x) = x + x^2 + 2*x^3 + 6*x^4 + 18*x^5 + 57*x^6 + 192*x^7 + 666*x^8 + 2362*x^9 + 8548*x^10 + ... + A273203(n)*x^n + ...
%e then R(x) and g.f. A(x) satisfy:
%e (1) R(A(x)) = x,
%e (2) R(x^2*A(x)) = x^3 - x^4.
%e GENERATING METHOD.
%e Define F(n), a polynomial in x of order 4^(n-1), by the following recurrence:
%e F(0) = x,
%e F(1) = (1 - x),
%e F(2) = (1 - x^3 * (1-x)),
%e F(3) = (1 - x^9 * (1-x)^3 * F(2)),
%e F(4) = (1 - x^27 * (1-x)^9 * F(2)^3 * F(3)),
%e F(5) = (1 - x^81 * (1-x)^27 * F(2)^9 * F(3)^3 * F(4)),
%e ...
%e F(n+1) = 1 - (1 - F(n))^3 * F(n)
%e ...
%e Then the g.f. A(x) equals the infinite product:
%e A(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...
%e that is,
%e A(x) = x * (1-x) * (1 - x^3*(1-x)) * (1 - x^9*(1-x)^3*(1 - x^3*(1-x))) * (1 - x^27*(1-x)^9*(1 - x^3*(1-x))^3*(1 - x^9*(1-x)^3*(1 - x^3*(1-x)))) * ...
%e SPECIFIC VALUES.
%e The infinite product formula allows us to evaluate the function A(x) at certain x rather quickly.
%e A(1/2) = (1/2) * (1/2) * (15/2^4) * (65521/2^16) * (18446744073488418241/2^64) * ... = 0.23432135581689318269048583...
%e A(2/3) = (2/3) * (1/3) * (73/3^4) * (43009345/3^16) * (3433683818046866504964267490561/3^64) * ... = 0.20010045690402248041761236...
%e A(1/3) = (1/3) * (2/3) * (79/3^4) * (43046089/3^16) * (3433683820292501618276703760129/3^64) * ... = 0.21673207172549998378147271...
%e The first relative maximum value of A(x) is given by
%e A(0.467248308756732593319037623545720...) = 0.235363377004699385202253369020528...
%o (PARI) /* Using Functional Equation in Definition */
%o {a(n) = my(A=[1,-1],B); for(i=1,n, A = concat(A,0);
%o R = serreverse(x*Ser(A));
%o A[#A] = polcoeff( x^3 - x^4 - subst(R,x,x^3*Ser(A)),#A+2) );H=A;A[n]}
%o for(n=1,60,print1(a(n),", "))
%o (PARI) /* Using Infinite Product Formula */
%o {F(n) = my(G=x); if(n==0,G=x, if(n==1,G=1-x, G = 1 - (1 - F(n-1))^3*F(n-1) ));G}
%o {a(n) = my(A = prod(k=0,round(log(n)/log(3))+1, F(k) +x*O(x^n))); polcoeff(A,n)}
%o for(n=1,60,print1(a(n),", "))
%Y Cf. A273203 (inverse), A273218, A350434.
%K sign
%O 1,5
%A _Paul D. Hanna_, Dec 30 2021
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