A324042 Number of triangular regions into which a figure made up of a row of n adjacent congruent rectangles is divided upon drawing diagonals of all possible rectangles.
4, 14, 32, 70, 124, 226, 360, 566, 820, 1218, 1696, 2310, 3020, 4018, 5160, 6590, 8196, 10218, 12464, 15110, 18012, 21650, 25624, 30142, 35028, 40954, 47344, 54558, 62284, 71034, 80360, 90806, 101892, 114770, 128416, 143286, 158972, 176914, 195816, 216350, 237908, 261546, 286304, 313102, 341100
Offset: 1
Keywords
Examples
For k adjacent congruent rectangles, the number of triangular regions in the j-th rectangle is: k\j| 1 2 3 4 5 6 7 ... ---+-------------------------------- 1 | 4, 0, 0, 0, 0, 0, 0, ... 2 | 7, 7, 0, 0, 0, 0, 0, ... 3 | 9, 14, 9, 0, 0, 0, 0, ... 4 | 11, 24, 24, 11, 0, 0, 0, ... 5 | 13, 30, 38, 30, 13, 0, 0, ... 6 | 15, 38, 60, 60, 38, 15, 0, ... 7 | 17, 44, 76, 86, 76, 44, 17, ... ... a(4) = 11 + 24 + 24 + 11 = 70.
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
- M. A. Alekseyev, M. Basova, and N. Yu. Zolotykh. On the minimal teaching sets of two-dimensional threshold functions. SIAM Journal on Discrete Mathematics 29:1 (2015), 157-165.
- Lars Blomberg, Scott R. Shannon, and N. J. A. Sloane, Graphical Enumeration and Stained Glass Windows, 1: Rectangular Grids, (2021). Also arXiv:2009.07918.
- Robert Israel, Maple program
- Jinyuan Wang, Illustration for n = 1, 2, 3, 4, 5
Programs
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Maple
V := proc(m,n,q) local a,i,j; a:=0; for i from 1 to m do for j from 1 to n do if gcd(i,j)=q then a:=a+(m+1-i)*(n+1-j); fi; od: od: a; end; a := n -> 2*( n*(n+1) + V(n,n,2) ); [seq(a(n), n=1..30)]; # N. J. A. Sloane, Mar 04 2020 See also Robert Israel link.
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Mathematica
Table[2 * (n^2 + n + Sum[Sum[Boole[GCD[i, j] == 2] * (n + 1 - i) * (n + 1 - j), {j, 1, n}], {i, 1, n}]), {n, 1, 45}] (* Joshua Oliver, Feb 05 2020 *) -
PARI
{ A324042(n) = 2*((n+1)*n + sum(i=1, n, sum(j=1, n, (gcd(i, j)==2)*(n+1-i)*(n+1-j))) ); } \\ Max Alekseyev, Jul 08 2019 -
Python
from sympy import totient def A324042(n): return 2*(2*n**2-n+1 + 2*sum(totient(i)*(n+1-2*i)*(n+1-i) for i in range(2,n//2+1))) # Chai Wah Wu, Aug 16 2021
Formula
a(n) = A177719(n+1) + 2*(n+1) = 2 * ( (n+1)*n + Sum_{i,j=1..n; gcd(i,j)=2} (n+1-i)*(n+1-j) ). - Max Alekseyev, Jul 08 2019
a(n) = 2*(2*n^2-n+1+2*Sum_{i=2..floor(n/2)} (n+1-2*i)*(n+1-i)*phi(i)). - Chai Wah Wu, Aug 16 2021
Extensions
a(8)-a(23) from Robert Israel, Jul 07 2019
Terms a(24) onward from Max Alekseyev, Jul 08 2019
Comments