| Displaying 1-6 of 6 results found.
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4, 30, 116, 290, 652, 1186, 2092, 3370, 5212, 7546, 10828, 14866, 20260, 26674, 34508, 43778, 55364, 68546, 84644, 102962, 124172, 147842, 175772, 206738, 242372, 281506, 325652, 374090, 429356, 488938, 556348, 629738, 710468, 797210, 892492, 994514, 1107668
COMMENTS
a(n)/ A331755(n) is the average number of polygons touching a non-boundary vertex in the graph defined in A306302.
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 vertices in the graph at which k polygons meet.
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7
4, 0, 1, 0, 4, 8, 0, 1, 0, 0, 28, 4, 2, 0, 1, 0, 0, 54, 4, 14, 0, 2, 0, 1, 0, 0, 124, 0, 22, 8, 2, 0, 2, 0, 1, 0, 0, 214, 0, 32, 4, 20, 0, 2, 0, 2, 0, 1, 0, 0, 382, 0, 50, 0, 26, 12, 2, 0, 2, 0, 2, 0, 1, 0, 0, 598, 0, 102, 0, 18, 4, 26, 0, 2, 0, 2, 0, 2, 0, 1
COMMENTS
For vertices not on the boundary, the number of polygons meeting at a vertex is simply the degree (or valency) of that vertex.
See A333275 for the degrees of the non-boundary vertices. Row n is the sum of [0, 0, ..., 0 (n-1 0's), 4, 2*n-2, 0, 0, ..., 0 (n 0's)] and row n of A333275.
EXAMPLE
Led d denote the number of polygons meeting at a vertex (except for boundary points, d is the degree of the vertex).
For n=2, the 4 corners have d=3, and on the center line there are 2 vertices with d=4 and 1 with d=6. In the interiors of each of the two squares there are 3 points with d=4.
So in total there are 4 points with d=3, 8 with d=4, and 1 with d=6. So row 2 of the triangle is [0, 4, 8, 0, 1].
The triangle begins:
4,0,1,
0,4,8,0,1,
0,0,28,4,2,0,1,
0,0,54,4,14,0,2,0,1,
0,0,124,0,22,8,2,0,2,0,1,
0,0,214,0,32,4,20,0,2,0,2,0,1;
0,0,382,0,50,0,26,12,2,0,2,0,2,0,1;
0,0,598,0,102,0,18,4,26,0,2,0,2,0,2,0,1;
0,0,950,0,126,0,32,0,30,16,2,0,2,0,2,0,2,0,1;
0,0,1334,0,198,0,62,0,20,4,32,0,2,0,2,0,2,0,2,0,1;
0,0,1912,0,286,0,100,0,10,0,34,20,2,0,2,0,2,0,2,0,2,0,1;
0,0,2622,0,390,0,118,0,38,0,22,4,38,0,2,0,2,0,2,0,2,0,2,0,1;
0,0,3624,0,510,0,136,0,74,0,10,0,38,24,2,0,2,0,2,0,2,0,2,0,2,0,1;
0,0,4690,0,742,0,154,0,118,0,10,0,24,4,44,0,2,0,2,0,2,0,2,0,2,0,2,0,1;
Consider the figure made up of a row of n adjacent congruent rectangles, with diagonals of all possible rectangles drawn; a(n) = number of interior vertices where exactly two lines cross.
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7
1, 6, 24, 54, 124, 214, 382, 598, 950, 1334, 1912, 2622, 3624, 4690, 6096, 7686, 9764, 12010, 14866, 18026, 21904, 25918, 30818, 36246, 42654, 49246, 57006, 65334, 75098, 85414, 97384, 110138, 124726, 139642, 156286, 174018, 194106, 214570, 237534, 261666, 288686, 316770, 348048, 380798, 416524, 452794, 492830
COMMENTS
It would be nice to have a formula or recurrence. - N. J. A. Sloane, Jun 22 2020
Irregular triangle read by rows: consider the regular n-gon defined in A007678. T(n,k) (n >= 1, k >= 4+2*t where t>=0) is the number of non-boundary vertices in the n-gon at which k polygons meet.
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6
0, 0, 0, 1, 5, 12, 1, 35, 40, 8, 1, 126, 140, 20, 0, 1, 330, 228, 60, 12, 0, 1, 715, 644, 112, 0, 0, 0, 1, 1365, 1168, 208, 0, 0, 0, 0, 1, 2380, 1512, 216, 54, 54, 0, 0, 0, 1, 3876, 3360, 480, 0, 0, 0, 0, 0, 0, 1, 5985, 5280, 660, 0, 0, 0, 0, 0, 0, 0, 1, 8855, 6144, 864, 264, 24, 0, 0, 0, 0, 0, 0, 12, 12650
FORMULA
If n = 2t+1 is odd then the n-th row has a single term, T(2t+1, 2t+4) = binomial(2t+1,4) (these values are given in A053126).
EXAMPLE
Table begins:
0;
0;
0;
1;
5;
12, 1;
35;
40, 8, 1;
126;
140, 20, 0, 1;
330;
228, 60, 12, 0, 1;
715;
644, 112, 0, 0, 0, 1;
1365;
1168, 208, 0, 0, 0, 0, 1;
2380;
1512, 216, 54, 54, 0, 0, 0, 1;
3876;
3360, 480, 0, 0, 0, 0, 0, 0, 1;
5985;
5280, 660, 0, 0, 0, 0, 0, 0, 0, 1;
8855;
6144, 864, 264, 24, 0, 0, 0, 0, 0, 0, 1;
12650;
11284, 1196, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
17550;
15680, 1568, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
23751;
13800, 2250, 420, 180, 120, 30, 0, 0, 0, 0, 0, 0, 0, 1;
31465;
28448, 2464, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
40920;
37264, 2992, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
52360;
12, 50, 152, 346, 732, 1294, 2232, 3546, 5428, 7806, 11136, 15226, 20676, 27150, 35048, 44386, 56044, 69302, 85480, 103882, 125180, 148942, 176968, 208034, 243772, 283014, 327272, 375826, 431212, 490918, 558456, 631978, 712844, 799726, 895152, 997322, 1110628
COMMENTS
a(n)/ A331755(n) is the average number of polygons touching a vertex in the graph defined in A306302.
a(n) = (n/4)*(n^3+2*n^2+5*n+8).
+10
4
0, 4, 17, 51, 124, 260, 489, 847, 1376, 2124, 3145, 4499, 6252, 8476, 11249, 14655, 18784, 23732, 29601, 36499, 44540, 53844, 64537, 76751, 90624, 106300, 123929, 143667, 165676, 190124, 217185, 247039, 279872, 315876, 355249, 398195, 444924, 495652, 550601, 609999, 674080, 743084, 817257, 896851, 982124, 1073340
COMMENTS
Consider a figure made up of a row of n >= 1 adjacent congruent rectangles in which all possible diagonals of the rectangles have been drawn. The number of regions formed is A306302. If we distort all these diagonals very slightly so that no three lines meet at a point, the number of regions changes to a(n).
FORMULA
Satisfies the identity a(n) = A306302(n) + Sum_{k=3..(n+1)} binomial(k-1,2)* A333275(n,2*k). E.g. for n=4 we have a(4) = 104 + 8*1 + 2*3 + 1*6 = 124. G.f.: x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
(End)
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 4, 17, 51, 124}, 50] (* or *)
PROG
(PARI) concat(0, Vec(x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5 + O(x^40))) \\ Colin Barker, May 27 2020
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