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A184630
Floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.
1
1, 6, 18, 43, 83, 144, 228, 341, 486, 666, 887, 1152, 1464, 1829, 2250, 2730, 3275, 3888, 4572, 5333, 6174, 7098, 8111, 9216, 10416, 11717, 13122, 14634, 16259, 18000, 19860, 21845, 23958, 26202, 28583, 31104, 33768, 36581, 39546, 42666, 45947, 49392, 53004, 56789, 60750, 64890, 69215, 73728, 78432, 83333, 88434
OFFSET
1,2
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
Cf. A184536.
Sequence in context: A272250 A272700 A191101 * A009957 A344992 A011929
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved