Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #24 Feb 20 2017 23:11:38
%S 2,6,13,24,37,54,73,96,121,150,181,216,253,294,337,384,433,486,541,
%T 600,661,726,793,864,937,1014,1093,1176,1261,1350,1441,1536,1633,1734,
%U 1837,1944,2053,2166,2281,2400,2521,2646,2773,2904,3037,3174,3313,3456,3601
%N a(n) = floor(1/{(2+n^3)^(1/3)}), where {}=fractional part.
%C Column 2 of the array at A184532.
%H G. C. Greubel, <a href="/A184533/b184533.txt">Table of n, a(n) for n = 1..5000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F a(n) = floor[1/{(2+n^3)^(1/3)}], where {}=fractional part.
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
%F a(n) = (6*n^2 - (1-(-1)^n))/4 for n>1.
%F From _Alexander R. Povolotsky_, Aug 22 2011: (Start)
%F a(n+1) +a(n) = 3*n^2 + 3*n + 1.
%F G.f. (-2 - 2*x - x^2 - 2*x^3 + x^4)/((-1 + x)^3*(1 + x)). (End)
%F a(n) = floor(1/((n^3+2)^(1/3)-n)). - _Charles R Greathouse IV_, Aug 23 2011
%t p[n_]:=FractionalPart[(n^3+2)^(1/3)]; q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,120}]
%t Join[{2},Table[(6*n^2 - (1-(-1)^n))/4,{n,2,50}]] (* or *) Join[{2}, LinearRecurrence[{2,0,-2,1},{6, 13, 24, 37},50]] (* _G. C. Greubel_, Feb 20 2017 *)
%o (PARI) a(n)=my(x=sqrtn(n^3+2,3));x-=n;1/x\1 \\ _Charles R Greathouse IV_, Aug 23 2011
%o (PARI) concat([2], for(n=2,25, print1((6*n^2 - (1-(-1)^n))/4, ", "))) \\ _G. C. Greubel_, Feb 20 2017
%Y Cf. A183532, A183534, A032528.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Jan 16 2011