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 A184533 a(n) = floor(1/{(2+n^3)^(1/3)}), where {}=fractional part. 3
 2, 6, 13, 24, 37, 54, 73, 96, 121, 150, 181, 216, 253, 294, 337, 384, 433, 486, 541, 600, 661, 726, 793, 864, 937, 1014, 1093, 1176, 1261, 1350, 1441, 1536, 1633, 1734, 1837, 1944, 2053, 2166, 2281, 2400, 2521, 2646, 2773, 2904, 3037, 3174, 3313, 3456, 3601 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Column 2 of the array at A184532. LINKS G. C. Greubel, Table of n, a(n) for n = 1..5000 Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). FORMULA a(n) = floor[1/{(2+n^3)^(1/3)}], where {}=fractional part. a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). a(n) = (6*n^2 - (1-(-1)^n))/4 for n>1. From Alexander R. Povolotsky, Aug 22 2011: (Start) a(n+1) +a(n) = 3*n^2 + 3*n + 1. G.f. (-2 - 2*x - x^2 - 2*x^3 + x^4)/((-1 + x)^3*(1 + x)). (End) a(n) = floor(1/((n^3+2)^(1/3)-n)). - Charles R Greathouse IV, Aug 23 2011 MATHEMATICA p[n_]:=FractionalPart[(n^3+2)^(1/3)]; q[n_]:=Floor[1/p[n]]; Table[q[n], {n, 1, 120}] 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 *) PROG (PARI) a(n)=my(x=sqrtn(n^3+2, 3)); x-=n; 1/x\1 \\ Charles R Greathouse IV, Aug 23 2011 (PARI) concat([2], for(n=2, 25, print1((6*n^2 - (1-(-1)^n))/4, ", "))) \\ G. C. Greubel, Feb 20 2017 CROSSREFS Cf. A183532, A183534, A032528. Sequence in context: A143689 A180773 A011891 * A338991 A178532 A003600 Adjacent sequences:  A184530 A184531 A184532 * A184534 A184535 A184536 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jan 16 2011 STATUS approved

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Last modified July 30 02:44 EDT 2021. Contains 346347 sequences. (Running on oeis4.)