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%I #14 Mar 21 2017 10:47:17
%S 1,6,18,43,83,144,228,341,486,666,887,1152,1464,1829,2250,2730,3275,
%T 3888,4572,5333,6174,7098,8111,9216,10416,11717,13122,14634,16259,
%U 18000,19860,21845,23958,26202,28583,31104,33768,36581,39546,42666,45947,49392,53004,56789,60750,64890,69215,73728,78432,83333,88434
%N Floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.
%H Ray Chandler, <a href="/A184630/b184630.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 2, -3, 3, -1).
%F a(n)=floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.
%F 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.
%t p[n_]:=FractionalPart[(n^4+6)^(1/4)]; q[n_]:=Floor[1/p[n]];
%t Table[q[n], {n, 1, 80}]
%t FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
%t Join[{1,6,18,43},LinearRecurrence[{3,-3,2,-3,3,-1},{83,144,228,341,486,666},47]] (* _Ray Chandler_, Aug 02 2015 *)
%Y Cf. A184536.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jan 18 2011