OFFSET
1,1
COMMENTS
Is a(n) = A001105(n) for n>=2 ?
LINKS
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).
FORMULA
a(n)=floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3) for n>=5, and that a(n)=2*n^2 for n>=2.
MATHEMATICA
p[n_]:=FractionalPart[(n^4+2*n)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{3}, LinearRecurrence[{3, -3, 1}, {8, 18, 32}, 69]] (* Ray Chandler, Aug 02 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved