login
A204686
G.f.: A(x) = Sum_{n>=0} x^n * A(x)^A006068(n) where A006068 forms the inverse permutation of the binary Gray code numbers (A003188).
0
1, 1, 2, 6, 18, 62, 228, 869, 3410, 13663, 55689, 230276, 963851, 4076429, 17394641, 74798044, 323797908, 1409980003, 6171824159, 27141256911, 119854448404, 531261779875, 2362873352568, 10541865261557, 47165485163426, 211572830443060, 951338664998601, 4287174496933723
OFFSET
0,3
COMMENTS
A006068 satisfies: A006068(n) XOR [A006068(n)/2] = n.
EXAMPLE
G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 18*x^4 + 62*x^5 + 228*x^6 +...
The g.f. A(x) satisfies:
A(x) = 1 + x*A(x) + x^2*A(x)^3 + x^3*A(x)^2 + x^4*A(x)^7 + x^5*A(x)^6 + x^6*A(x)^4 + x^7*A(x)^5 + x^8*A(x)^15 + x^9*A(x)^14 + x^10*A(x)^12 +...
where the powers of A(x) are given by A006068, which begins:
[0,1,3,2,7,6,4,5,15,14,12,13,8,9,11,10,31,30,28,29,24,25,27,26,...].
PROG
(PARI) {A006068(n)=local(B=n); for(k=1, floor(log(n+1)/log(2)), B=bitxor(B, n\2^k)); B}
{a(n)=local(A=1+x); for(i=1, n, A=sum(m=0, n, x^m*(A+x*O(x^n))^A006068(m))); polcoeff(A, n)}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 18 2012
STATUS
approved