OFFSET
0,2
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).
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)
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Luce ETIENNE, Dec 17 2017
STATUS
approved