|
%I #7 Mar 30 2012 18:37:34
%S 1,1,2,6,18,62,228,869,3410,13663,55689,230276,963851,4076429,
%T 17394641,74798044,323797908,1409980003,6171824159,27141256911,
%U 119854448404,531261779875,2362873352568,10541865261557,47165485163426,211572830443060,951338664998601,4287174496933723
%N 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).
%C A006068 satisfies: A006068(n) XOR [A006068(n)/2] = n.
%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 18*x^4 + 62*x^5 + 228*x^6 +...
%e The g.f. A(x) satisfies:
%e 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 +...
%e where the powers of A(x) are given by A006068, which begins:
%e [0,1,3,2,7,6,4,5,15,14,12,13,8,9,11,10,31,30,28,29,24,25,27,26,...].
%o (PARI) {A006068(n)=local(B=n);for(k=1,floor(log(n+1)/log(2)),B=bitxor(B,n\2^k));B}
%o {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)}
%Y Cf. A192483, A006068, A003188.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 18 2012
|
