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%I #5 Jan 26 2014 15:37:05
%S 1,3,4,14,21,53,69,237,321,867,1044,3638,5441,13667,17684,60854,81921,
%T 221187,266244,931854,1397781,3495477,4542789,15555437,21053441
%N Results from a change in the rules leading to sequence A097357.
%C Let b_n(i) be defined as for sequence A097357. Then A097357(n) = sum(i=0...infty)b_n(i) = sum(i=1...n)b_n(i). We have: Stage 1: (1,0,0,0,0,0,0,0,0,0,0,0,0) = b_1 (disregarding initial 0); Stage 2: (1,1,0,0,0,0,0,0,0,0,0,0,0,) = b_2 (disregarding initial 0); Stage 3: (0,0,1,0,0,0,0,0,0,0,0,0,0,) = b_3 (disregarding initial 0); Stage 4: (0,1,1,1,0,0,0,0,0,0,0,0,0,) = b_4 (disregarding initial 0); Stage 5: (1,0,1,0,1,0,0,0,0,0,0,0,0,) = b_5 (disregarding initial 0); Stage 6: (1,0,1,0,1,1,0,0,0,0,0,0,0,) Stage 7: (1,0,1,0,0,0,1,0,0,0,0,0,0,) Stage 8: (1,0,1,1,0,1,1,1,0,0,0,0,0,) a(n) is defined as the number which results from interpreting the sequence b_n as a binary string read backwards from the first nonzero term.
%e a(6) = 53 since b_6 = (1,0,1,0,1,1,0,0,0,0,0,0,0) and 110101 written in base 10 is 53.
%Y Cf. A097357.
%K nonn
%O 1,2
%A _Creighton Dement_, Sep 12 2005
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