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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 16 2011
STATUS
approved