A176646 a(n) is the number of convex pentagons in an n-triangular net.

0, 0, 3, 21, 78, 216, 498, 1014, 1884, 3264, 5349, 8379, 12642, 18480, 26292, 36540, 49752, 66528, 87543, 113553, 145398, 184008, 230406, 285714, 351156, 428064, 517881, 622167, 742602, 880992, 1039272, 1219512, 1423920, 1654848, 1914795, 2206413, 2532510, 2896056

Offset: 1

Author

Jonathan Vos Post, Dec 21 2010

Keywords

• nonn
• easy

Comments

See P(n) in Theorem 2.1, p.2 of Zhu.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Jun-Ming Zhu, The number of convex pentagons and hexagons in an n-triangular net, arXiv:1012.4058 [math.CO], 2010. See P(n), formula (1), page 2.
• Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1).

Crossrefs

Cf. A166189 (for the hexagons).

Programs

Magma

[(1/320)*(12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n): n in [1..40]]; // G. C. Greubel, Jul 02 2021

Maple

A176646:= n-> (12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n)/320;
seq(A176646(n), n=1..40); # R. J. Mathar, Dec 21 2010

Mathematica

LinearRecurrence[{5,-9,5,5,-9,5,-1}, {0,0,3,21,78,216,498}, 40] (* Harvey P. Dale, Jan 14 2015 *)

PARI

f(k) = (12*k^5 + 25*k^4 + 5*k^3 - 10*k^2 - 2*k)/10;
g(k) = (12*k^5 - 5*k^4 - 15*k^3 + 5*k^2 + 3*k)/10;
a(n) = if (n%2, f((n-1)/2), g(n/2)); \\ Michel Marcus, Jul 04 2021

Sage

[(1/320)*(12*n^5 -10*n^4 -60*n^3 +40*n^2 +48*n -15 +15*(-1)^n) for n in (1..40)] # G. C. Greubel, Jul 02 2021

Formula

From G. C. Greubel, Jul 03 2021: (Start)

a(n) = (1/320)*(12*n^5 - 10*n^4 - 60*n^3 + 40*n^2 + 48*n - 15 + 15*(-1)^n).

a(2*n+1) = n*(n+1)*(12*n^3 + 13*n^2 - 8*n - 2)/10.

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

G.f.: 3*x^3*(1 + 2*x)/((1 + x)*(1 - x)^6).

E.g.f.: (1/320)*(15*exp(-x) - (15 -30*x +30*x^2 -180*x^3 -110*x^4 -12*x^5)*exp(x)). (End)

a(n) = 3*A001753(n-3) + 6*A001753(n-4). - R. J. Mathar, Jul 04 2021

Extensions

Definition corrected and edited by Michel Marcus and G. C. Greubel, Jul 03 2021