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A380996 a(n) is the number of vertices in the n-fold iterated barycentric subdivision of a triangle (or 2-simplex).

%I A380996 #19 Mar 10 2025 18:07:50
%S A380996 3,7,25,121,673,3937,23425,140161,840193,5039617,30234625,181401601,
%T A380996 1088397313,6530359297,39182106625,235092541441,1410555052033,
%U A380996 8463329918977,50779978727425,304679870791681,1828079221604353,10968475323334657,65810851927425025,394865111539384321
%N A380996 a(n) is the number of vertices in the n-fold iterated barycentric subdivision of a triangle (or 2-simplex).
%C A380996 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.
%C A380996 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.
%H A380996 Wikipedia, <a href="https://en.wikipedia.org/wiki/Barycentric_subdivision">Barycentric subdivision</a>.
%H A380996 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-20,12).
%F A380996 a(n) = (6^n + 3*2^n)/2 + 1.
%F A380996 From _Stefano Spezia_, Feb 23 2025: (Start)
%F A380996 G.f.: (3 - 20*x + 22*x^2)/((1 - x)*(1 - 2*x)*(1 - 6*x)).
%F A380996 E.g.f.: exp(x)*(2 + 3*exp(x) + exp(5*x))/2. (End)
%F A380996 a(n) = 3*A074502(n-1) - 2 for n>0. - _Hugo Pfoertner_, Feb 23 2025
%e A380996 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.
%t A380996 Table[(6^n + 3*2^n)/2 + 1, {n, 0, 50}]
%o A380996 (Python)
%o A380996 def A380996(n): return 3**n+3<<n-1|1 if n else 3 # _Chai Wah Wu_, Mar 10 2025
%Y A380996 Cf. A000400, A007283, A074502.
%K A380996 easy,nonn
%O A380996 0,1
%A A380996 _Mikhail Lavrov_, Feb 11 2025