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%I #10 Jan 26 2016 14:29:02
%S 3,5,11,3,21,3,5,43,3,5,11,3,85,3,5,11,3,21,3,5,171,3,5,11,3,21,3,5,
%T 43,3,5,11,3,341,3,5,11,3,21,3,5,43,3,5,11,3,85,3,5,11,3,21,3,5,683,3,
%U 5,11,3,21,3,5,43,3,5
%N A268032 with repeated 1's removed.
%C Records appear to be given by A001045 Jacobsthal numbers.
%C (a(n)-1)/2 appears to be A085358.
%C The terms between the A001045(n+3) are:
%C 3
%C 5
%C 11
%C 3,
%C 21
%C 3, 5,
%C 43
%C 3, 5, 11, 3,
%C 85
%C 3, 5, 11, 3, 21, 3, 5,
%C 171
%C 3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3,
%C 341
%C 3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3, 85, 3, 5, 11, 3, 21, 3, 5,
%C 683
%C This gives the same sequence. Every column has the same number.
%C By rows,there are 0, 0, 1, 2, 4, 7, 12, 20, ... apparently = Fib(n+1) - 1 = A000071 terms.(Comment from Paul Curtz).
%C From Paul Curtz, Jan 26 2016: (start)
%C a(n) is also in
%C 0, 1, 1 0, 3, 0, 1, 5, 0, ... equivalent to A035614(n)
%C 1, 1, 3, 1, 5, 1, 1, 11, 1, ... equivalent to A035612(n)
%C 1, 3, 5, 1, 11, 1, 3, 21, 1, ... (compare to A268032)
%C 3, 5, 11, 3, 21, 3, 5, 43, 3, ... a(n) (equivalent to a3(n) in A035612)
%C 5, 11, 21, 5, 43, 5, 11, 85, 5, ...
%C etc.
%C Every vertical comes from A001045 (*).
%C Second row: first one removing all 0's.
%C Third row: second one removing a part of 1's respecting (*)
%C Fourth row: third one removing all 1's.
%C etc.
%C The offset 0 is homogeneous to these sequences. (End)
%e A268032 begins 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 3, 1, 1, 1, 21, ... hence this sequence begins 3, 5, 11, 3, 21, ...
%Y Cf. A000071, A001045, A035612, A035614, A085358, A268032.
%K nonn
%O 1,1
%A _Jeremy Gardiner_, Jan 24 2016
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