Original entry on oeis.org
0, 1, 8, 7, 24, 22, 40, 22, 80, 54, 80, 22, 80, 54, 112, 70, 240, 134, 144, 22, 80, 54, 112, 70, 240, 134, 176, 70, 240, 150, 336, 246, 720, 326, 272, 22, 80, 54, 112, 70, 240, 134, 176, 70, 240, 150, 336, 246, 720, 326, 304, 70, 240, 150, 336, 246, 720, 342, 464, 246, 752, 454, 1136, 838, 2064
Offset: 0
Written as an irregular triangle the sequence begins:
0;
1;
8;
7;
24;
22, 40;
22, 80, 54, 80;
22, 80, 54, 112, 70, 240, 134, 144;
22, 80, 54, 112, 70, 240, 134, 176, 70, 240, 150, 336, 246, 720, 326, 272;
22, 80, 54, 112, 70, 240, 134, 176, 70, 240, 150, 336, 246, 720, 326, 304, 70,...
...
Starting from a(3) = 7 the row lengths of triangle are the terms of A011782.
A187214
Number of gulls (or G-toothpicks) added at n-th stage in the first quadrant of the gullwing structure of A187212.
Original entry on oeis.org
0, 1, 1, 2, 2, 4, 5, 4, 2, 4, 6, 6, 8, 14, 15, 8, 2, 4, 6, 6, 8, 14, 16, 10, 8, 14, 18, 20, 30, 44, 39, 16, 2, 4, 6, 6, 8, 14, 16, 10, 8, 14, 18, 20, 30, 44, 40, 18, 8, 14, 18, 20, 30, 44, 42, 28, 30, 46, 56, 70, 104, 128, 95
Offset: 1
At stage 1 we start in the first quadrant from a Q-toothpick centered at (1,0) with its endpoints at (0,0) and (1,1). There are no gulls in the structure, so a(1) = 0.
At stage 2 we place a gull (or G-toothpick) with its midpoint at (1,1) and its endpoints at (2,0) and (2,2), so a(2) = 1. There is only one exposed midpoint at (2,2).
At stage 3 we place a gull with its midpoint at (2,2), so a(3) = 1. There are two exposed endpoints.
At stage 4 we place two gulls, so a(4) = 2. There are two exposed endpoints.
At stage 5 we place two gulls, so a(5) = 2. There are four exposed endpoints.
And so on.
If written as a triangle begins:
0,
1,
1,2,
2,4,5,4,
2,4,6,6,8,14,15,8,
2,4,6,6,8,14,16,10,8,14,18,20,30,44,39,16,
2,4,6,6,8,14,16,10,8,14,18,20,30,44,40,18,8,14,18,20,30,44,42,28,...
It appears that rows converge to A151688.
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