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A184536 a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.

%I A184536 #36 Jun 13 2025 10:58:36
%S A184536 5,32,108,256,500,864,1372,2048,2916,4000,5324,6912,8788,10976,13500,
%T A184536 16384,19652,23328,27436,32000,37044,42592,48668,55296,62500,70304,
%U A184536 78732,87808,97556,108000,119164,131072,143748,157216,171500,186624,202612,219488,237276,256000,275684,296352,318028,340736,364500,389344,415292,442368,470596,500000,530604,562432,595508,629856,665500,702464,740772,780448,821516,864000,907924,953312,1000188
%N A184536 a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.
%C A184536 For n >= 1, the value of (n^4+1)^(1/4) is just slightly above n, so the fractional part is (n^4+1)^(1/4)-n. For n > 1, then, 4*n^3 < 1/((1+n^4)^(1/4)-n) < 4*n^3+1. [Proof that 4*n^3*((1+n^4)^(1/4)-n) < 1 follows easily by isolating the quartic root and raising to the 4th power; similarly, 1 < (4*n^3+1)*((1+n^4)^(1/4)-n) needs a sign estimation of an 8th-order polynomial.] In conclusion, a(n)=A033430(n) for n > 1. - _Bruno Berselli_, Jan 30 2011
%H A184536 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A184536 a(n) = floor[1/{(1+n^4)^(1/4)}], where {}=fractional part.
%F A184536 G.f.: x*(x^4-4*x^3+10*x^2+12*x+5)/(x-1)^4. - _Colin Barker_, Sep 21 2012
%t A184536 p[n_]:=FractionalPart[(n^4+1)^(1/4)];
%t A184536 q[n_]:=Floor[1/p[n]];
%t A184536 Table[q[n],{n,1,80}]
%t A184536 LinearRecurrence[{4,-6,4,-1},{5,32,108,256,500},70] (* _Harvey P. Dale_, Dec 14 2023 *)
%Y A184536 Cf. A184537. Essentially the same as A033430.
%K A184536 nonn,easy
%O A184536 1,1
%A A184536 _Clark Kimberling_, Jan 16 2011