A184538 Floor[1/{(3+n^4)^(1/4)}], where {}=fractional part.

2, 11, 36, 85, 166, 288, 457, 682, 972, 1333, 1774, 2304, 2929, 3658, 4500, 5461, 6550, 7776, 9145, 10666, 12348, 14197, 16222, 18432, 20833, 23434, 26244, 29269, 32518, 36000, 39721, 43690, 47916, 52405, 57166, 62208, 67537, 73162, 79092, 85333, 91894, 98784, 106009, 113578, 121500, 129781, 138430, 147456, 156865, 166666, 176868, 187477, 198502, 209952, 221833, 234154, 246924, 260149, 273838, 288000

Offset: 1

Author

Clark Kimberling, Jan 16 2011

Keywords

• nonn

Crossrefs

A184536.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+3)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n],{n,1,80}]

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]