A257093 a(n) = n*(n+1)*(13*n+2)/6.

0, 5, 28, 82, 180, 335, 560, 868, 1272, 1785, 2420, 3190, 4108, 5187, 6440, 7880, 9520, 11373, 13452, 15770, 18340, 21175, 24288, 27692, 31400, 35425, 39780, 44478, 49532, 54955, 60760, 66960, 73568, 80597, 88060, 95970, 104340, 113183, 122512, 132340

Offset: 0

Author

Luce ETIENNE, Apr 16 2015

Keywords

• nonn
• easy

Comments

This sequence gives the number of triangles of all sizes in (5*n^2)-polyiamonds in a tetragonal or hexagonal or heptagonal configuration.

It is the sum of (1/2)*Sum_{j=0..n-1} (n-j)*(5*n+1-j) triangles oriented in one direction and (1/2)*Sum_{j-0..n-1} (n-j)*(5*n-1-3*j) oriented in the opposite direction.

Shäfli's notation: 3.3.3.3.3 for a(1).

The difference between this sequence and A050409(n) equals A000292(n-1).

Also, (1/3)*(A002717(2*n) + A255211(n) - 2*A000330(n)) gives A033994(n): a (5*n^2)-polyiamond in pentagonal configuration that does not belong to this sequence because a(1)=6.

a(n) is odd only when n mod 4 = 1.

Examples

			Second comment a(0) = 0; a(1) = 3 + 2; a(2) = 16 + 12; a(3) = 46 + 36; a(4) = 100 + 80; a(5) = 185 + 150; a(6) = 308 + 252.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A002411, A011379, A033429, A050409, A000292, A002717, A000330, A033994, A255211.

Programs

Magma

[n*(n+1)*(13*n+2)/6: n in [0..40]]; // Vincenzo Librandi, Apr 16 2015

Mathematica

Table[n (n + 1) (13 n + 2)/6, {n, 0, 40}] (* Vincenzo Librandi, Apr 16 2015 *)
CoefficientList[Series[x (5+8x)/(1-x)^4,{x,0,50}],x] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,28,82},60] (* Harvey P. Dale, Feb 12 2023 *)

Formula

a(n) = Sum_{j=0..n-1} (n-j)*(5*n-2*j).

From Vincenzo Librandi, Apr 16 2015: (Start)

G.f.: x*(5+8*x)/(1-x)^4.

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

E.g.f.: exp(x)*x*(30 + 54*x + 13*x^2)/6. - Stefano Spezia, Mar 02 2025

Extensions

Corrected by Harvey P. Dale, Feb 12 2023