%I A120880
%S A120880 1,2,1,2,4,2,1,2,1,2,4,2,4,8,4,2,4,2,1,2,1,2,4,2,1,2,1,2,4,2,4,8,4,2,4,
%T A120880 2,4,8,4,8,16,8,4,8,4,2,4,2,4,8,4,2,4,2,1,2,1,2,4,2,1,2,1,2,4,2,4,8,4,
               2,
%U A120880 4,2,1,2,1,2,4,2,1,2,1,2,4,2,4,8,4,2,4,2,4,8,4,8,16,8,4,8,4,2,4,2,4,8,
               4
%N A120880 G.f. satisfies: A(x) = A(x^3)*(1 + 2*x + x^2); thus a(n) = 2^A062756(n), 
               where A062756(n) is the number of 1's in the ternary expansion of 
               n.
%C A120880 More generally, if g.f. of {a(n)} satisfies: A(x) = A(x^3)*(1 + b*x + 
               c*x^2), then a(n) = b^A062756(n)*c^A081603(n), where A062756(n) is 
               the number of 1's and A081603(n) is the number of 2's, in the ternary 
               expansion of n. This sequence is not the same as A059151.
%C A120880 a(n) is the number of entries in the n-th row of Pascal's triangle that 
               are congruent to 1 mod 3 minus the number of entries that are congruent 
               to 2 mod 3. - Naoki Sato (nsato7(AT)yahoo.ca), Jun 22 2007
%F A120880 a((3^n+1)/2) = 2^n; a(n) = a(floor(n/3))*2^[[n (mod 3)] (mod 2)], with 
               a(0)=1. G.f.: A(x) = prod_{n>=0} (1 + x^(3^n))^2. Self-convolution 
               of A039966. Row sums of triangle A117947(n,k) = balanced ternary 
               of C(n,k) mod 3.
%e A120880 Records are 2^n at positions: 0,1,4,13,40,121,...,(3^n-1)/2,... (n>=0).
%e A120880 A(x) = 1 + 2*x + x^2 + 2*x^3 + 4*x^4 + 2*x^5 + x^6 + 2*x^7 + x^8 +...
%o A120880 (PARI) a(n)=if(n==0,1,a(n\3)*2^((n%3)%2))
%Y A120880 Cf. A117947, A039966, A062756, A081603.
%Y A120880 Sequence in context: A121439 A009205 A086754 this_sequence A059151 A059149 
               A013943
%Y A120880 Adjacent sequences: A120877 A120878 A120879 this_sequence A120881 A120882 
               A120883
%K A120880 nonn
%O A120880 0,2
%A A120880 Paul D. Hanna (pauldhanna(AT)juno.com), Jul 11 2006