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

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

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: A180773 A011891 A370880 * A338991 A178532 A003600

Adjacent sequences: A184530 A184531 A184532 * A184534 A184535 A184536

Keyword

nonn,easy

Author

Clark Kimberling, Jan 16 2011

Status

approved