A366932 Total number of edges formed after n points have been placed in general position on each edge of a triangle (as in A365929).

3, 9, 51, 255, 855, 2193, 4719, 8991, 15675, 25545, 39483, 58479, 83631, 116145, 157335, 208623, 271539, 347721, 438915, 546975, 673863, 821649, 992511, 1188735, 1412715, 1666953, 1954059, 2276751, 2637855, 3040305, 3487143, 3981519, 4526691, 5126025, 5782995, 6501183, 7284279, 8136081, 9060495, 10061535, 11143323, 12310089, 13566171

Offset

0,1

Comments

See A365929 for more information. See A365929 and A367015 for images of the triangle.

Links

Paolo Xausa, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

Conjecture: a(n) = 3*(3*n^4/2 - 2*n^3 + 3*n^2/2 + n + 1). [This is now a theorem - N. J. A. Sloane, Dec 31 2025]

a(n) = A367015(n) + 3*A366478(n) - 1 by Euler's formula.

From Elmo R. Oliveira, Apr 17 2026: (Start)

G.f.: 3*(1 + x)*(1 - 3*x + 15*x^2 + 5*x^3)/(1 - x)^5.

E.g.f.: 3*exp(x)*(2 + 4*x + 12*x^2 + 14*x^3 + 3*x^4)/2.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)

Maple

a366932:= n -> 3*(3*n^4/2 - 2*n^3 + 3*n^2/2 + n + 1); # N. J. A. Sloane, Jan 01 2026

Mathematica

A366932[n_] := 3*(n*(3*n + 2) + 1)*(n*(n - 2) + 2)/2; Array[A366932, 50, 0] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {3, 9, 51, 255, 855}, 50] (* Paolo Xausa, Jan 02 2026 *)

Crossrefs

Cf. A365929 (internal vertices), A367015 (regions), A366478.

Equals 3*A389626.

Sequence in context: A079836 A171951 A323232 * A319105 A106329 A193521

Adjacent sequences: A366929 A366930 A366931 * A366933 A366934 A366935

Keyword

nonn,easy

Author

Scott R. Shannon, Nov 02 2023

Extensions

Definition edited by N. J. A. Sloane, Dec 31 2025

Status

approved