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%I #15 May 15 2017 03:07:18
%S 1,3,7,12,18,27,36,48,60,75,90,108,126,147,168,192,216,243,270,300,
%T 330,363,396,432,468,507,546,588,630,675,720,768,816,867,918,972,1026,
%U 1083,1140,1200,1260,1323,1386,1452,1518,1587,1656,1728,1800,1875,1950,2028,2106,2187,2268,2352,2436,2523,2610,2700,2790,2883,2976,3072,3168,3267,3366,3468,3570,3675,3780,3888,3996,4107,4218,4332,4446,4563,4680,4800,4920,5043,5166,5292,5418,5547,5676,5808
%N a(n) = floor(1/{(4+n^3)^(1/3)}), where {}=fractional part.
%H G. C. Greubel, <a href="/A184534/b184534.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = floor[1/{(4+n^3)^(1/3)}], where {}=fractional part.
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
%F From _Colin Barker_, Oct 07 2012: (Start)
%F Empirical: a(n) = 3*(1 - (-1)^n + 4*n + 2*n^2)/8 for n>2.
%F Empirical G.f.: x*(x^6-2*x^5+x^4-x^2-x-1)/((x-1)^3*(x+1)).(End)
%t Table[Floor[1/FractionalPart[(n^3 + 4)^(1/3)]], {n, 1, 120}]
%o (PARI) for(n=1, 50, print1(floor(1/frac((4 + n^3)^(1/3))), ", ")) \\ _G. C. Greubel_, May 14 2017
%Y Cf. A183532, A183534.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jan 16 2011
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