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%I #4 Mar 30 2012 18:36:38
%S 1,1,3,1,5,1,1,11,1,1,3,1,21,1,1,3,1,5,1,1,43,1,1,3,1,5,1,1,11,1,1,3,
%T 1,85,1,1,3,1,5,1,1,11,1,1,3,1,21,1,1,3,1,5,1,1,171,1,1,3,1,5,1,1,11,
%U 1,1,3,1,21,1,1,3,1,5,1,1,43,1,1,3,1,5,1,1,11,1,1,3,1,341,1,1,3,1,5,1,1,11
%N Runs of zeros in binomial(3k+2,k+1)/(3k+2) modulo 2 (A085405).
%C Construction: start with strings S(1)={1}, S(2)={1,3}, S(3)={1,5}; to obtain S(k) for k>3, concatenate all previous strings excluding S(k-1), then replace the last number L in this resulting string with {4L-(-1)^k}. This sequence is the limit of S(k) as k grows, generating strings with Fibonacci growth: {1,1,11}, {1,1,3,1,21}, {1,1,3,1,5,1,1,43}, {1,1,3,1,5,1,1,11,1,1,3,1,85}, ...
%F For n>0, a(F(n))=A001045(n), where F(n) is the n-th Fibonacci number with F(1)=1, F(2)=2.
%e To generate string S(4) at k=4: concatenate {S(1),S(2)} = {1, 1,3}, then replace the last number L=3 by 4*L-1=11 to obtain S(4)={1,1,11}.
%e At k=5: concatenate {S(1),S(2),S(3)} = {1, 1,3, 1,5}, then replace the last number L=5 by 4*L+1=21 to obtain S(5)={1,1,3,1,21}.
%Y Cf. A006013, A001045, A085405, A022340 (positions of ones).
%K nonn
%O 1,3
%A _Paul D. Hanna_, Jun 29 2003
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