OFFSET
1,1
FORMULA
a(n)=floor[1/{(3+n^4)^(1/4)}], where {}=fractional part.
Recurrence relation appears to be a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 3*a(n-4) + 3*a(n-5) - a(n-6).
Empirical G.f.: x*(x+1)*(x^6-3*x^5+3*x^4+6*x^2+3*x+2)/((x-1)^4*(x^2+x+1)). [Colin Barker, Sep 21 2012]
MATHEMATICA
p[n_]:=FractionalPart[(n^4+3)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 16 2011
STATUS
approved