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A296636
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Sequences n*(n+1)*(6*n+1)/2 and n*(n+1)*(7*n+1)/2 interleaved.
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1
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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
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OFFSET
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0,2
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COMMENTS
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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].
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LINKS
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FORMULA
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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.
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)
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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