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A184538
Floor[1/{(3+n^4)^(1/4)}], where {}=fractional part.
0
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
1,1
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