cp's OEIS Frontend

A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

%I A184633 #25 Mar 21 2017 10:48:39
%S A184633 1,4,12,28,55,96,152,227,324,444,591,768,976,1219,1500,1820,2183,2592,
%T A184633 3048,3555,4116,4732,5407,6144,6944,7811,8748,9756,10839,12000,13240,
%U A184633 14563,15972,17468,19055,20736,22512,24387,26364,28444,30631,32928,35336,37859,40500,43260,46143,49152,52288,55555,58956,62492,66167,69984,73944,78051,82308,86716,91279,96000,100880,105923,111132,116508,122055,127776,133672,139747,146004,152444,159071,165888,172896,180099,187500
%N A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.
%H A184633 Edward Jiang and Ray Chandler, <a href="/A184633/b184633.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Edward Jiang)
%H A184633 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 2, -3, 3, -1).
%F A184633 a(n) = floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.
%F A184633 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>=9.
%F A184633 Empirical g.f.: x*(x+1)*(x^6-3*x^5+3*x^4-x^3+3*x^2+1) / ((x-1)^4*(x^2+x+1)). - _Colin Barker_, Jun 13 2015
%t A184633 p[n_]:=FractionalPart[(n^4+9)^(1/4)]; q[n_]:=Floor[1/p[n]];
%t A184633   Table[q[n], {n, 1, 80}]
%t A184633   FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
%t A184633 Join[{1,4},LinearRecurrence[{3,-3,2,-3,3,-1},{12,28,55,96,152,227},73]] (* _Ray Chandler_, Aug 02 2015 *)
%o A184633 (PARI) a(n)=my(t=(9+n^4)^(1/4)); 1\(t-t\1) \\ _Charles R Greathouse IV_, Sep 12 2014
%Y A184633 Cf. A184536.
%K A184633 nonn
%O A184633 1,2
%A A184633 _Clark Kimberling_, Jan 18 2011