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A184637 a(n) = floor(1/{(n^4+3*n)^(1/4)}), where {}=fractional part.

%I A184637 #20 Sep 08 2022 08:45:55
%S A184637 2,6,12,21,33,48,65,85,108,133,161,192,225,261,300,341,385,432,481,
%T A184637 533,588,645,705,768,833,901,972,1045,1121,1200,1281,1365,1452,1541,
%U A184637 1633,1728,1825,1925,2028,2133,2241,2352,2465,2581,2700,2821,2945,3072,3201,3333,3468,3605,3745,3888,4033,4181,4332,4485,4641,4800,4961,5125,5292,5461,5633,5808,5985,6165,6348,6533,6721,6912,7105,7301,7500,7701,7905,8112,8321,8533
%N A184637 a(n) = floor(1/{(n^4+3*n)^(1/4)}), where {}=fractional part.
%H A184637 G. C. Greubel, <a href="/A184637/b184637.txt">Table of n, a(n) for n = 1..10000</a>
%H A184637 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,1).
%F A184637 a(n) = 2*a(n-1) -a(n-2) +a(n-3) -2*a(n-4) +a(n-5) for n>=8.
%F A184637 G.f.: x*(x+1)*(x^5-2*x^4+x^3-2*x^2-2)/((x-1)^3*(x^2+x+1)). - _Colin Barker_, Oct 07 2012
%F A184637 a(n) = n^2 + floor(n^2/3) with n>2, a(1)=2, a(2)=6. This confirms the g.f. and the recurrence. - _Bruno Berselli_, Aug 08 2013
%t A184637 p[n_]:=FractionalPart[(n^4+3*n)^(1/4)];
%t A184637 q[n_]:=Floor[1/p[n]];
%t A184637 Table[q[n], {n, 1, 80}]
%t A184637 FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
%t A184637 Join[{2,6},LinearRecurrence[{2,-1,1,-2,1},{12,21,33,48,65},78]] (* _Ray Chandler_, Aug 02 2015 *)
%o A184637 (Magma) Fp:=func<i | Root(i^4+3*i,4)-Iroot(i^4+3*i,4)>; [Floor(1/Fp(n)): n in [1..80]]; // _Bruno Berselli_, Aug 08 2013
%o A184637 (PARI) x='x+O('x^30); Vec(x*(x+1)*(x^5-2*x^4+x^3-2*x^2-2)/((x-1)^3*(x^2 +x+1))) \\ _G. C. Greubel_, Apr 04 2018
%Y A184637 Cf. A184535, A184636.
%K A184637 nonn,easy
%O A184637 1,1
%A A184637 _Clark Kimberling_, Jan 18 2011