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A333275
Irregular triangle read by rows: consider the graph defined in A306302 formed from a row of n adjacent congruent rectangles by drawing the diagonals of all visible rectangles; T(n,k) (n >= 1, 2 <= k <= 2n+2) is the number of non-boundary vertices in the graph at which k polygons meet.
10
0, 0, 1, 0, 0, 6, 0, 1, 0, 0, 24, 0, 2, 0, 1, 0, 0, 54, 0, 8, 0, 2, 0, 1, 0, 0, 124, 0, 18, 0, 2, 0, 2, 0, 1, 0, 0, 214, 0, 32, 0, 10, 0, 2, 0, 2, 0, 1, 0, 0, 382, 0, 50, 0, 22, 0, 2, 0, 2, 0, 2, 0, 1, 0, 0, 598, 0, 102, 0, 18, 0, 12, 0, 2, 0, 2, 0, 2, 0, 1
OFFSET
1,6
COMMENTS
The number of polygons meeting at a non-boundary vertex is simply the degree (or valency) of that vertex.
Row sums are A159065.
Sum_k k*T(n,k) gives A333277.
See A333274 for the degrees if the boundary vertices are included.
T(n,k) = 0 if k is odd. But the triangle includes those zero entries because this is used to construct A333274.
FORMULA
It seems that T(n, 2*k) = A119437(n+1, k) / 2 unless k = n+1. - Andrei Zabolotskii, Jun 06 2026
EXAMPLE
Led d denote the number of polygons meeting at a vertex.
For n=2, in the interiors of each of the two squares there are 3 points with d=4, and the center point has d=6.
So in total there are 6 points with d=4 and 1 with d=6. So row 2 of the triangle is [0, 0, 6, 0, 1].
The triangle begins:
0,0,1,
0,0,6,0,1,
0,0,24,0,2,0,1,
0,0,54,0,8,0,2,0,1,
0,0,124,0,18,0,2,0,2,0,1,
0,0,214,0,32,0,10,0,2,0,2,0,1,
0,0,382,0,50,0,22,0,2,0,2,0,2,0,1,
0,0,598,0,102,0,18,0,12,0,2,0,2,0,2,0,1
...
If we leave out the uninteresting zeros, the triangle begins:
[1]
[6, 1]
[24, 2, 1]
[54, 8, 2, 1]
[124, 18, 2, 2, 1]
[214, 32, 10, 2, 2, 1]
[382, 50, 22, 2, 2, 2, 1]
[598, 102, 18, 12, 2, 2, 2, 1]
[950, 126, 32, 26, 2, 2, 2, 2, 1]
[1334, 198, 62, 20, 14, 2, 2, 2, 2, 1]
[1912, 286, 100, 10, 30, 2, 2, 2, 2, 2, 1]
[2622, 390, 118, 38, 22, 16, 2, 2, 2, 2, 2, 1]
... - N. J. A. Sloane, Jul 27 2020
KEYWORD
nonn,tabf,changed
AUTHOR
EXTENSIONS
a(36) and beyond from Lars Blomberg, Jun 17 2020
STATUS
approved