OFFSET
1,2
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).
FORMULA
a(n)=floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=11, which implies a(n) = (2*n^3-1+A049347(n))/3 for n>=5.
MATHEMATICA
p[n_]:=FractionalPart[(n^4+6)^(1/4)]; q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 6, 18, 43}, LinearRecurrence[{3, -3, 2, -3, 3, -1}, {83, 144, 228, 341, 486, 666}, 47]] (* Ray Chandler, Aug 02 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved