A248851 a(n) = ( 2*n*(2*n^2 + 9*n + 14) + (-1)^n - 1 )/16.

0, 3, 10, 22, 41, 68, 105, 153, 214, 289, 380, 488, 615, 762, 931, 1123, 1340, 1583, 1854, 2154, 2485, 2848, 3245, 3677, 4146, 4653, 5200, 5788, 6419, 7094, 7815, 8583, 9400, 10267, 11186, 12158, 13185, 14268, 15409, 16609, 17870, 19193, 20580, 22032

Offset

0,2

Comments

Consider a grid of small triangles of side 1 forming a regular polygon with side n*(n+2); a(n) is the number of equilateral triangles of side length >= 1 in this figure which are oriented with the sides of the figure.

This sequence gives the number of triangles of all sizes in a (n^2+2*n)-iamond with a 4*n-gon configuration.

Equals (1/2)*Sum_{j=0..n-1} (n-j)*(n+1-j) + (-1 + (1/8)*Sum_{j=0..(2*n+1-(-1)^n)/4} (2*n+3+(-1)^n-4*j)*(2*n+3-(-1)^n-4*j)) numbers of triangles in a direction and in the opposite direction.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Formula

G.f.: x*(x^3-2*x^2+x+3) / ((x-1)^4*(x+1)). - Colin Barker, Mar 03 2015

a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5). - Colin Barker, Mar 03 2015

Example

From third comment: a(0)=0, a(1)=1+2, a(2)=4+6, a(3)=10+12, a(4)=20+21, a(5)=35+33.

Mathematica

CoefficientList[Series[x (x^3 - 2 x^2 + x + 3) / ((x - 1)^4(x + 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 21 2015 *)
LinearRecurrence[{3, -2, -2, 3, -1}, {0, 3, 10, 22, 41}, 50] (* Harvey P. Dale, Jan 17 2023 *)

Prog

(PARI) concat(0, Vec(x*(x^3-2*x^2+x+3)/((x-1)^4*(x+1)) + O(x^100))) \\ Colin Barker, Mar 03 2015
(Magma) [(4*n^3+18*n^2+28*n-(1-(-1)^n)) div 16: n in [0..50]]; // Vincenzo Librandi, Mar 21 2015

Crossrefs

Cf. A117143, A000292, A002623, A045947, A005563, A008586.

Sequence in context: A122795 A140066 A006503 * A023554 A294414 A299336

Adjacent sequences: A248848 A248849 A248850 * A248852 A248853 A248854

Keyword

nonn,easy

Author

Luce ETIENNE, Mar 03 2015

Extensions

Typo in formula fixed by Vincenzo Librandi, Mar 21 2015

Name rewritten using the closed form by Bruno Berselli, Apr 19 2015

Status

approved