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A101190 G.f.: A(x) = Sum_{n>=0} a(n)/2^A005187(n) * x^n = lim_{n->oo} F(n)^(1/2^n) where F(n) is defined by F(n) = F(n-1)^2 + x^(2^n-1) for n >= 1 with F(0) = 1.

%I A101190 #10 Mar 07 2024 05:08:40
%S A101190 1,1,-1,5,-53,127,-677,2221,-61133,205563,-1394207,4852339,-68586849,
%T A101190 243751723,-1741612525,6265913725,-363239625661,1323861506899,
%U A101190 -9699189175227,35700526467479,-527987675255931,1960112858076289,-14606721595781139,54604708004873403
%N A101190 G.f.: A(x) = Sum_{n>=0} a(n)/2^A005187(n) * x^n = lim_{n->oo} F(n)^(1/2^n) where F(n) is defined by F(n) = F(n-1)^2 + x^(2^n-1) for n >= 1 with F(0) = 1.
%F A101190 G.f. A(x) = ( Sum_{n>=0} A101191(n)/2^A004134(n) * x^n )^2.
%F A101190 G.f. A(x) satisfies A(2*x)^2 = Sum_{n>=0} A101189(n)*(2*x)^n.
%e A101190 G.f.: A(x) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 + ... + a(n)/2^A005187(n)*x^n + ...
%e A101190 where 2^A005187(n) is also the denominator of [x^n] 1/sqrt(1-x).
%e A101190 GENERATING METHOD.
%e A101190 We can illustrate the generating method for g.f. A(x) as follows.
%e A101190 Given F(n) = F(n-1)^2 + (2*x)^(2^n-1) for n >= 1 with F(0) = 1,
%e A101190 the first few polynomials generated by F(n) begin
%e A101190 F(0) = 1,
%e A101190 F(1) = F(0)^2 + x^(2^1-1) = 1 + x,
%e A101190 F(2) = F(1)^2 + x^(2^2-1) = 1 + 2*x + x^2 + x^3,
%e A101190 F(3) = F(2)^2 + x^(2^3-1) = 1 + 4*x + 6*x^2 + 6*x^3 + 5*x^4 + 2*x^5 + x^6 + x^7.
%e A101190 ...
%e A101190 The 2^n-th roots of F(n) tend to the limit of the g.f.:
%e A101190 F(1)^(1/2^1) = 1 + 1/2*x - 1/8*x^2 + 1/16*x^3 - 5/128*x^4 + 7/256*x^5 - 21/1024*x^6 + 33/2048*x^7 - 429/32768*x^8 + ...
%e A101190 F(2)^(1/2^2) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 1965/2048*x^7 - 46797/32768*x^8 + ...
%e A101190 F(3)^(1/2^3) = 1 + 1/2*x - 1/8*x^2 + 5/16*x^3 - 53/128*x^4 + 127/256*x^5 - 677/1024*x^6 + 2221/2048*x^7 - 61133/32768*x^8 + ...
%e A101190 ...
%e A101190 The limit of this process equals the g.f. A(x) of this sequence.
%e A101190 Note: the sum of the coefficients in F(n) equals A003095(n):
%e A101190 1, 2 = 1 + 1, 5 = 1 + 2 + 1 + 1, 26 = 1 + 4 + 6 + 6 + 5 + 2 + 1 + 1, ...
%e A101190 The last n coefficients in F(n) read backwards are Catalan numbers (A000108).
%e A101190 POWERS OF A(x).
%e A101190 The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:
%e A101190 A^(2^0) = [1, 1/2, -1/8, 5/16, -53/128, 127/256, -677/1024, 2221/2048, ...],
%e A101190 A^(2^1) = [1, 1, 0, 1/2, -1/2, 1/2, -5/8, 9/8, -2, 53/16, -89/16, 155/16, ...],
%e A101190 A^(2^2) = [1, 2, 1, 1, 0, 0, 0, 1/2, -1, 3/2, -5/2, 9/2, -8, 14, -197/8, 44, ...],
%e A101190 A^(2^3) = [1, 4, 6, 6, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1/2, -2, 5, ...],
%e A101190 A^(2^4) = [1, 8, 28, 60, 94, 116, 114, 94, 69, 44, 26, 14, 5, 2, 1, 1, 0, 0, ...].
%o A101190 (PARI) {a(n) = my(F=1,A,L); if(n==0,A=1,L=ceil(log(n+1)/log(2)); for(k=1,L, F = F^2 + x^(2^k-1) +x*O(x^n)); A = polcoeff(F^(1/2^L),n)); numerator(A)}
%o A101190 for(n=0,32, print1(a(n),", "))
%Y A101190 Cf. A101189, A101191, A005187, A003095, A000108.
%K A101190 sign
%O A101190 0,4
%A A101190 _Paul D. Hanna_, Dec 03 2004
%E A101190 Entry revised by _Paul D. Hanna_, Mar 05 2024