A299965 Number of triangles in a Star of David of size n.

0, 20, 118, 354, 788, 1480, 2490, 3878, 5704, 8028, 10910, 14410, 18588, 23504, 29218, 35790, 43280, 51748, 61254, 71858, 83620, 96600, 110858, 126454, 143448, 161900, 181870, 203418, 226604, 251488, 278130, 306590, 336928, 369204, 403478, 439810, 478260, 518888

Offset: 0

Author

John King, Feb 22 2018

Keywords

• nonn
• easy
• less

Comments

In a Star of David of size n, there are A135453(n) "size=1" triangles.

The number of matchstick units is A045946.

Examples

			For n=1, there are 12 (size=1) + 6 (size=4) + 2 (size=9) = 20 triangles.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000 (corrected and extended original b-file from Colin Barker after data change).
• John King, Star a=6, 84 matches, 118 triangles
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A045950, A045946, A135453,

For the total number of triangles in a different arrangement, see A002717 (for triangular matchstick), A045949 (for hexagonal matchstick).

Programs

Mathematica

A299965[n_] := n*(n*(10*n + 9) + 1); Array[A299965, 50, 0] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {0, 20, 118, 354}, 50] (* Paolo Xausa, Sep 18 2024 *)

PARI

concat(0, Vec(2*x*(10 + 19*x + x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Apr 04 2019

Formula

a(n) = n*(10*n^2+9*n+1) = 2*A045950(n).

From Colin Barker, Apr 04 2019: (Start)

G.f.: 2*x*(10 + 19*x + x^2) / (1 - x)^4.

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

E.g.f.: exp(x)*x*(20 + 39*x + 10*x^2). - Stefano Spezia, Sep 20 2024

Extensions

Corrected by John King, Stefano Spezia, and Paolo Xausa, Sep 20 2024