A265282 Number of triangles in a certain geometric structure: see "Illustration of initial terms" link for precise definition.

0, 1, 3, 5, 10, 13, 22, 26, 41, 46, 68, 74, 105, 112, 153, 161, 214, 223, 289, 299, 380, 391, 488, 500, 615, 628, 762, 776, 931, 946, 1123, 1139, 1340, 1357, 1583, 1601, 1854, 1873, 2154, 2174, 2485, 2506, 2848, 2870, 3245, 3268, 3677, 3701, 4146, 4171, 4653

Offset: 0

Author

Luce ETIENNE, Dec 06 2015

Keywords

• nonn
• easy

Comments

In words: This sequence gives the number of triangles of all sizes in a (2*n^2+8*n-1+(-1)^n)/8-polyiamond with a (7*n-2-(n-2)*(-1)^n)/4-gon: we have (2*n^3+9*n^2+31*n+21+3*(n^2-5*n-7)*(-1)^n)/96 triangles in a direction and (2*n^3+27*n^2+109*n-66+3*(n^2+9*n+18)*(-1)^n+12*(-1)^((2*n-1+(-1)^n)/4))/192 triangles in the other direction. (But the Illustration link is far more informative. - N. J. A. Sloane, Jan 23 2016)

At stage n, we count (2*n^2 + 6*n + 3 - (2*n+3)*(-1)^n)/16 triangles of size 1 in one direction and (2*n^2 + 10*n - 5 + (2*n+5)*(-1)^n)/16 triangles of size 1 in the opposite direction. The total number of triangles of size 1 in both directions is A024206(n+1).

We observe that a(4)=10 strengthens the Pythagorean relation between 4 and 10 (Tetraktys): cf. triangular numbers, A000217; and that it is from n = 4 we can see and count hexagonal and dodecagonal forms, for example, in a reticular system (incomplete with hexagonal holes) by opposition to the compact shape obtained from A002717.

We can obtain this reticular system from A248851.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000
• Luce ETIENNE, Illustration of initial terms
• Luce ETIENNE, A265282 from A248851
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,0,0,-2,2,1,-1).

Crossrefs

Cf. A000217, A000292, A001477, A001859, A002620, A002717, A004006, A004526, A045947, A132337.

Cf. A248851.

Programs

Magma

[(2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n + 4*(-1)^((2*n - 1 + (-1)^n) div 4)) / 64: n in [0..50]]; // Vincenzo Librandi, Dec 07 2015

Mathematica

Table[(2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n +
    4*(-1)^((2*n - 1 + (-1)^n)/4))/64, {n, 0, 100}] (* G. C. Greubel, Dec 20 2015 *)
LinearRecurrence[{1,2,-2,0,0,-2,2,1,-1},{0,1,3,5,10,13,22,26,41},60] (* Harvey P. Dale, Aug 07 2019 *)

PARI

vector(100, n, n--; (2*n^3+15*n^2+57*n-8+(3*n^2-n+4)*(-1)^n+4*(-1)^((2*n-1+(-1)^n)/4))/64) \\ Altug Alkan, Dec 06 2015

PARI

concat(0, Vec(x*(1+2*x+x^3-x^4-x^5+x^7)/((1-x)^4*(1+x)^3*(1+x^2)) + O(x^100))) \\ Colin Barker, Dec 07 2015

Formula

a(n) = A045947(floor(n/2)) + A024206(n+1). Note that A045947(floor(n/2)) = (2*n^3-n^2-7*n+(3*n^2-n-4)*(-1)^n+4*(-1)^((2*n-1+(-1)^n)/4))/64.

a(n) = (2*n^3 + 15*n^2 + 57*n - 8 + (3*n^2 - n + 4)*(-1)^n + 4*(-1)^((2*n - 1 + (-1)^n)/4))/64.

G.f.: x*(1+2*x+x^3-x^4-x^5+x^7) / ((1-x)^4*(1+x)^3*(1+x^2)). - Colin Barker, Dec 07 2015

Extensions

a(26) corrected by Altug Alkan, Dec 06 2015