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

Author

Clark Kimberling, Jan 16 2011

Keywords

• nonn
• easy

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).

Crossrefs

Cf. A183532, A183534. Essenitally the same as A032528.

Programs

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 *)

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

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.: x*(-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

E.g.f.: (3*x*(x + 1)*cosh(x) + (3*x^2 + 3*x - 1)*sinh(x) + 2*x)/2. - Stefano Spezia, Apr 19 2025