A101191 - OEIS

%I #4 Mar 30 2012 18:36:44

%S 1,1,-3,23,-525,2695,-29687,191991,-10488701,70977675,-968279181,

%T 6752850945,-191225421641,1363019302883,-19538003443615,

%U 140961586090743,-16379289413266717,119621607825995891,-1755802638936696081,12944528671963135869,-383361262914445548739

%N G.f.: A(x) = Sum_{n>=0}a(n)/2^A004134(n)*x^n = limit_{n->oo} F(n)^(1/2^(n+1)) where F(n) is the n-th iteration of: F(0) = 1, F(n) = F(n-1)^2 + x^(2^n-1) for n>=1.

%C Although the power series for the g.f. A(x) diverges at x=1, the Euler transform of the power series A(x) at x=1 converges to the constant A076949: Sum_{n>=0}[Sum_{k=0..n}C(n,k)*a(k))/2^A004134(n) ]/2^(n+1) = 1.2259024435...

%F G.f. A(x) satisfies: A(x)^2 = Sum_{n>=0} A101190(n)/2^A005187(n)*x^n. G.f. A(x) satisfies: A(2*x)^4 = Sum_{n>=0} A101189(n)*(2x)^n.

%e The iteration begins:

%e F(0) = 1,

%e F(1) = F(0)^2 + x^(2^1-1) = 1 +x,

%e F(2) = F(1)^2 + x^(2^2-1) = 1 +2*x +x^2 +x^3,

%e 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 The 2^(n+1)-th roots of F(n) tend to the limit of the g.f.:

%e F(1)^(1/2^2) = 1 +1/4*x -3/32*x^2 +7/128*x^3 -77/2048*x^4 +231/8192*x^5 +...

%e F(2)^(1/2^3) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...

%e F(3)^(1/2^4) = 1 +1/4*x -3/32*x^2 +23/128*x^3 -525/2048*x^4 +2695/8192*x^5 +...

%e The limit of this process is the g.f. A(x) of this sequence.

%e The coefficients of x^k in the 2^n powers of the g.f. A(x) begin:

%e A^(2^0)=[1,1/4,-3/32,23/128,-525/2048,2695/8192,-29687/65536,...],

%e A^(2^1)=[1,1/2,-1/8,5/16,-53/128,127/256,-677/1024,2221/2048,...],

%e A^(2^2)=[1,1,0,1/2,-1/2,1/2,-5/8,9/8,-2,53/16,-89/16,155/16,...],

%e A^(2^3)=[1,2,1,1,0,0,0,1/2,-1,3/2,-5/2,9/2,-8,14,-197/8,44,...],

%e A^(2^4)=[1,4,6,6,5,2,1,1,0,0,0,0,0,0,0,1/2,-2,5,...],

%e A^(2^5)=[1,8,28,60,94,116,114,94,69,44,26,14,5,2,1,1,0,0,...].

%e Note: the sum of the coefficients of x^k in F(n) equals A003095(n+1):

%e 1, 2=1+1, 5=1+2+1+1, 26=1+4+6+6+5+2+1+1, ...

%e The last n coefficients in F(n) read backwards are Catalan numbers (A000108).

%o (PARI) {a(n)=local(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)); A=polcoeff(F^(1/2^(L+1))+x*O(x^n),n));numerator(A)}

%Y Cf. A101189, A101190, A005187, A003095, A076949.

%K sign

%O 0,3

%A _Paul D. Hanna_, Dec 03 2004