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 A184533 floor(1/{(2+n^3)^(1/3)}), where {}=fractional part. 2

%I

%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 floor(1/{(2+n^3)^(1/3)}), where {}=fractional part.

%C Column 2 of the array at A184532.

%F a(n)=floor[1/{(2+n^3)^(1/3)}], where {}=fractional part.

%F Recurrence appears to be a(n)=2a(n-1)-2a(n-3)+a(n-4), equivalent to a(n) = 3*n^2/2 -(1-(-1)^n)/4 for n>1.

%F Equivalent a(n+1) +a(n) = 3*n^2+3*n+1 and G.f. (-2 - 2*x - x^2 - 2*x^3 + x^4)/((-1 + x)^3*(1 + x)) [Alexander R. Povolotsky, Aug 22 2011]

%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)];

%t q[n_]:=Floor[1/p[n]];

%t Table[q[n],{n,1,120}]

%o (PARI) a(n)=my(x=sqrtn(n^3+2,3));x-=n;1/x\1 \\ _Charles R Greathouse IV_, Aug 23 2011

%Y Cf. A183532, A183534, A032528.

%K nonn,easy,changed

%O 1,1

%A _Clark Kimberling_, Jan 16 2011

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