A069497 Triangular numbers of the form 6*k.
0, 6, 36, 66, 78, 120, 210, 276, 300, 378, 528, 630, 666, 780, 990, 1128, 1176, 1326, 1596, 1770, 1830, 2016, 2346, 2556, 2628, 2850, 3240, 3486, 3570, 3828, 4278, 4560, 4656, 4950, 5460, 5778, 5886, 6216, 6786, 7140, 7260, 7626, 8256, 8646, 8778, 9180
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).
Programs
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Maple
a[0] := 0:a[1] := 3:a[2] := 8:a[3] := 11:seq((12*(floor(i/4))+a[i mod 4])*(12*(floor(i/4))+a[i mod 4]+1)/2,i=0..100);
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Mathematica
CoefficientList[ Series[ 6x (x^2 -x +1) (x^2 +4x +1)/((x^2 +1)^2*(1 -x)^3), {x, 0, 45}], x] (* or *) LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 6, 36, 66, 78, 120, 210}, 46] (* Robert G. Wilson v, May 31 2017 *) Select[Accumulate[Range[0, 89]], Divisible[#, 6] &] (* Alonso del Arte, May 31 2017 *)
Formula
a(n) = 6 * A154293(n). - Joerg Arndt, Aug 18 2022
From R. J. Mathar, Nov 18 2009: (Start)
a(n) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7).
G.f.: 6*x*(x^2-x+1)*(x^2+4*x+1)/((1+x^2)^2*(1-x)^3) (6*A154293). (End)
From Amiram Eldar, Aug 18 2022: (Start)
Sum_{n>=2} 1/a(n) = 2 - (3+4*sqrt(3))*Pi/18. (End)
Extensions
More terms from Sascha Kurz, Apr 01 2002
More terms from R. J. Mathar, Aug 21 2007
Offset corrected to 1, Joerg Arndt, Aug 18 2022