OFFSET
0,1
COMMENTS
The barycentric subdivision of a triangle divides it into six smaller triangles by drawing the three medians; new vertices are added at the midpoints of the sides and at the centroid, where the medians intersect. To iterate this process, take the barycentric subdivision of all the triangles created in the process.
Viewed as a planar graph, the n-fold barycentric subdivision has 6^n triangles and an unbounded face of length 3*2^n. Taking half the sum of the face lengths, we conclude that there are (3*6^n + 3*2^n)/2 edges. Finally, we can find the number of vertices by Euler's formula, V-E+F=2.
LINKS
Wikipedia, Barycentric subdivision.
Index entries for linear recurrences with constant coefficients, signature (9,-20,12).
FORMULA
a(n) = (6^n + 3*2^n)/2 + 1.
From Stefano Spezia, Feb 23 2025: (Start)
G.f.: (3 - 20*x + 22*x^2)/((1 - x)*(1 - 2*x)*(1 - 6*x)).
E.g.f.: exp(x)*(2 + 3*exp(x) + exp(5*x))/2. (End)
a(n) = 3*A074502(n-1) - 2 for n>0. - Hugo Pfoertner, Feb 23 2025
EXAMPLE
a(0)=1 because a triangle has 3 vertices. After one barycentric subdivision, we add 3 vertices at the midpoints of the sides, and 1 vertex at the center, for a(1)=7 vertices total. A second barycentric subdivision adds 12 more midpoints and 6 more centroids, so a(2)=25.
MATHEMATICA
Table[(6^n + 3*2^n)/2 + 1, {n, 0, 50}]
PROG
(Python)
def A380996(n): return 3**n+3<<n-1|1 if n else 3 # Chai Wah Wu, Mar 10 2025
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Mikhail Lavrov, Feb 11 2025
STATUS
approved