A296636 Sequences n*(n+1)*(6*n+1)/2 and n*(n+1)*(7*n+1)/2 interleaved.

0, 7, 8, 39, 45, 114, 132, 250, 290, 465, 540, 777, 903, 1204, 1400, 1764, 2052, 2475, 2880, 3355, 3905, 4422, 5148, 5694, 6630, 7189, 8372, 8925, 10395, 10920, 12720, 13192, 15368, 15759, 18360, 18639, 21717, 21850, 25460, 25410, 29610, 29337, 34188, 33649, 39215, 38364, 44712

Offset: 0

Author

Luce ETIENNE, Dec 17 2017

Keywords

• nonn
• easy

Comments

Difference between these subsequences is A002411.

This sequence gives numbers of triangles all sizes in every n-th stage [of what? - N. J. A. Sloane, Feb 09 2018].

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Luce ETIENNE, Illustration
• Eric Weisstein's World of Mathematics, Polyiamond
• Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Crossrefs

Cf. A002411, A002717, A006331, A033581, A033582, A255211.

Programs

GAP

List([0..50], n -> (2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128); # Bruno Berselli, Feb 12 2018

Magma

[(2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128: n in [0..50]]; // Bruno Berselli, Feb 12 2018

Mathematica

CoefficientList[Series[x (7 + 8 x + 11 x^2 + 13 x^3)/((1 - x)^4*(1 + x)^4), {x, 0, 46}], x] (* Michael De Vlieger, Dec 18 2017 *)
LinearRecurrence[{0,4,0,-6,0,4,0,-1},{0,7,8,39,45,114,132,250},50] (* Harvey P. Dale, May 01 2018 *)
Rest[Flatten[Table[With[{c=(n(n+1))/2},{c*(6n+1),c*(7n+1)}],{n,0,30}]]] (* Harvey P. Dale, Oct 11 2020 *)

PARI

concat(0, Vec(x*(7 + 8*x + 11*x^2 + 13*x^3) / ((1 - x)^4*(1 + x)^4) + O(x^80))) \\ Colin Barker, Dec 18 2017

Formula

a(n) = a(n-1)+4*a(n-2)-4*a(n-3)-6*a(n-4)+6*a(n-5)+4*a(n-6)-4*a(n-7)-a(n-8)+a(n-9).

a(n) = (2*n+1-(-1)^n)*(2*n+5-(-1)^n)*(13*n+10+(n-6)*(-1)^n)/128.

From Colin Barker, Dec 18 2017: (Start)

G.f.: x*(7 + 8*x + 11*x^2 + 13*x^3) / ((1 - x)^4*(1 + x)^4).

a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n>7.

(End)