OFFSET
0,2
FORMULA
Limit_{n->infinity} Sum_{k=0..n-1} [x^k] G(x)^(1/2^n) = 2, where G(x) satisfies [x^n] G(x)^(1/2^n) = 1 for n>=0.
EXAMPLE
This sequence forms the numerators of coefficients in G(x), which begin:
[1,2,11/2,247/16,87453/2048,30392377/262144,83081803051/268435456,...].
The denominators are 2^b(n) where b(n) takes on the values:
[0,0,1,4,11,18,28,39,55,70,88,107,130,153,179,206,239,270,304,339,...]
which is b(n) = n*(n-1) - A000120(n) for n>1 with b(0)=b(1)=0.
Illustrate [x^n] G(x)^(1/2^n) = 1 for n=0..5 by:
G(x) = (1)+2x +11x^2/2 +247x^3/2^4 +87453x^4/2^11 +30392377x^5/2^18 +...;
G(x)^(1/2) = 1 +(x)+9x^2/2^2 +175x^3/2^5 +54685x^4/2^12 +16941497x^5/2^19 +..;
G(x)^(1/4) = 1 +x/2 +(x^2)+143x^3/2^6 +41437x^4/2^13 +119466176x^5/2^20 +...;
G(x)^(1/8) = 1 +x/2^2 +15x^2/2^5 +(x^3)+35541x^4/2^14 +9826265x^5/2^21 +...;
G(x)^(1/16) = 1 +x/2^3 +29x^2/2^7 +483x^3/2^10 +(x^4) +8853753x^5/2^22 +...;
G(x)^(1/32) = 1 +x/2^4 +57x^2/2^9 +1875x^3/2^13 +251395x^4/2^19 +(x^5)+...;
so that the coefficient of x^n in G(x)^(1/2^n) equals 1 for n>=0.
To illustrate that the n-th partial sums of G(x)^(1/2^n) approaches 2: at n=5, Sum_{k=0..4} [x^k] G(x)^(1/32) = 1+1/2^4+57/2^9+1875/2^13+251395/2^19 = 1.88...
PROG
(PARI) {a(n)=local(A=[1]); if(n==1, 2, for(i=0, n, A=Vec(Ser(concat(Vec(Ser(A)^(1/2^#A)), 1))^(2^#A))); A[n+1]*2^(n*(n-1))/2^subst(Pol(binary(n)), x, 1))}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 26 2007, Oct 29 2007
STATUS
approved