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a(3*k) = (k+1)*(3*k^2+3*k+2)/2.
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Robert Israel, <a href="/A217330/b217330.txt">Table of n, a(n) for n = 0..10000</a>
<a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,2,-4,2,-1,2,-1).
From Robert Israel, May 09 2018: (Start)
a(3*k) = (k+1)*(3*k^2+3*k+2)/2.
a(3*k+1) = (k+1)*(3*k^2+6*k+4)/2.
a(3*k+2) = 3*(k+1)^2*(k+2)/2.
G.f.: (1+2*x^3)/((1-x)*(1-x^3))^2. (End)
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1, 2, 3, 8, 13, 18, 30, 42, 54, 76, 98, 120, 155, 190, 225, 276, 327, 378, 448, 518, 588, 680, 772, 864, 981, 1098, 1215, 1360, 1505, 1650, 1826, 2002, 2178, 2388, 2598, 2808, 3055, 3302, 3549, 3836, 4123, 4410, 4740, 5070, 5400, 5776, 6152, 6528, 6953, 7378, 7803
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The number of integer solutions to the equation x1 + x2 + x3 + x4 = n, with xi >= 0, and with x2 + x3 divisible by 3.
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