|
%I #5 Mar 30 2012 18:36:44
%S 1,1,2,2,5,2,5,5,12,2,5,5,12,5,12,12,30,2,5,5,12,5,12,12,30,5,12,12,
%T 30,12,30,30,73,2,5,5,12,5,12,12,30,5,12,12,30,12,30,30,73,5,12,12,30,
%U 12,30,30,73,12,30,30,73,30,73,73,169,2,5,5,12,5,12,12,30,5,12,12,30,12,30
%N A fractal sequence such that a(n) = A101911(A000120(n-1)) for n>0, where A101911 is the binomial transform of this sequence.
%C A101911 also forms the records in this sequence at positions 2^k for k>=0. A000120 is the binary 1's-counting sequence. Removing the even-indexed terms shifts this sequence one place left.
%F a(n) = Sum_{k=0, n-1} Mod(C(n-1, k), 2)*a(A000120(n-k-1)) for n>0, a(0)=1. a(2^k) = A101911(k) for k>=0.
%e Denote the n-th term of the binomial transform by: b(n)=A101911(n):
%e A101911 = {1,2,5,12,30,73,169,377,831,1842,4110,...}.
%e Note A000120 = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,...}.
%e Then this sequence is formed by the following construct:
%e {1,b(0),b(1),b(1),b(2),b(1),b(2),b(2),b(3),...,b(A000120(n-1)),...}
%e so that a(2^0)=b(0), a(2^1)=b(1), a(2^2)=b(2), a(2^3)=b(3), ...
%o (PARI) {a(n)=if(n==0,1,sum(k=0,n-1, (binomial(n-1,k)%2)*a(subst(Pol(binary(n-k-1)),x,1))))}
%Y Cf. A101911, A000120.
%K eigen,nonn
%O 0,3
%A _Paul D. Hanna_, Dec 21 2004
|
