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a(n) = 3n^2 - 3n + 2 for n > 0.

Proof (from Joshua Zucker and N. J. A. Sloane, Dec 01 2017)

Represent the configuration of n triangles by a planar graph with a node for each vertex of the triangles and for each intersection point. Let there be v_n nodes and e_n edges. By classical graph theory, a(n) = e_n - v_n + 2. When we go from n to n+1 triangles, each side of the new triangle can meet each side of the existing triangles at most twice, so Dv_n := v_{n+1}-v_n <= 6n.

Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1}-e_n = 3 + 2*Dv_n, Da_n := a(n+1)-a(n) = 3 + Dv_n <= 3+6*n.

These upper bounds can be achieved by taking 3n points equally spaced around a circle and drawing n concentric overlapping equilateral triangles in the obvious way, and we achieve a(n) = 3n^2 - 3n + 2 (and v_n = 3n^2, e_n = 3n(2n-1)) for n>0. QED

a(n) is the nearest integer to (Sum_{k>=n} 1/k^2)/(Sum_{k>=n} 1/k^4). - Benoit Cloitre, Jun 12 2003

a(n) = a(n-1) + 6*n - 6 (with a(1) = 2). - Vincenzo Librandi, Dec 07 2010

For n > 0, a(n) = A002061(n-1) + A056220(n); and for n > 1, a(n) = A002061(n+1) + A056220(n-1). - Bruce J. Nicholson, Sep 22 2017

A349508(n)/A349509(n) <= a(n) < A349511(n) < A349512(n) (see Corollary 7 in Zhang et al., 2021).

a(n) ~ (n/6)^(3*n*(n-1))*exp(-6+13/n+3*n^2)/(3*sqrt(6*Pi)).

From Chai Wah Wu, May 30 2016: (Start)

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.

G.f.: 2*x*(4*x^2 - 2*x + 1)/(x - 1)^4. (End)