0, 1, 2, 3, 5, 4, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 17, 16, 18, 20, 21, 19, 22, 24, 25, 28, 29, 23, 27, 26, 30, 31, 33, 32, 34, 36, 37, 35, 38, 40, 41, 44, 45, 39, 43, 42, 47, 50, 54, 58, 59, 55, 51, 46, 48, 49, 52, 53, 56, 60, 61, 57, 62, 63, 65, 64, 66, 68, 69, 67, 70, 72, 73, 76, 77, 71, 75, 74, 79, 82, 86, 90, 91, 87, 83, 78, 80, 81
Offset: 0
Please look at Paul Tek's illustration: We start at root, 0, go up to 1, visit its left child 2 (which is a leaf), before proceeding the infinite trunk (A179016) to 3, then visit first the leaf 5 at the right hand side, before proceeding the infinite trunk to 4, then visit the leaf 6 at the left hand side, before proceeding the infinite trunk right to 7, from which we first visit the leaf 9 at the right hand side, before proceeding the infinite trunk to 8 at the left hand side. Thus we have ten initial terms of the sequence: 0, 1, 2, 3, 5, 4, 6, 7, 9, 8, ...
From 8 we proceed first to the left 10, because it is not a part of the infinite trunk, and we traverse a finite side-tree ("tendril") of three nodes in order 10, 12, 13, only after which we proceed the infinite trunk to the right, to 11, thus we have the next four terms of the sequence 10, 12, 13, 11.
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