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A184634 a(n) = floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.

%I A184634 #19 Feb 25 2019 16:44:12
%S A184634 1,3,11,25,50,86,137,204,291,400,532,691,878,1097,1350,1638,1965,2332,
%T A184634 2743,3200,3704,4259,4866,5529,6250,7030,7873,8780,9755,10800,11916,
%U A184634 13107,14374,15721,17150,18662,20261,21948,23727,25600,27568,29635,31802,34073,36450,38934,41529,44236,47059,50000,53060,56243,59550,62985,66550,70246,74077,78044,82151,86400,90792,95331,100018,104857,109850,114998,120305,125772,131403,137200
%N A184634 a(n) = floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.
%H A184634 Robert Israel, <a href="/A184634/b184634.txt">Table of n, a(n) for n = 1..10000</a>
%H A184634 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1, 0, 1, -3, 3, -1).
%F A184634 a(n)=floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.
%F A184634 It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8).
%F A184634 From _Robert Israel_, Feb 25 2019: (Start)
%F A184634 2*n^3/5 + 3/(2*n) > 1/{(10+n^4)^(1/4)} > 2*n^3/5 for all n.
%F A184634 From this we can show that a(5*k) = 50*k^3 for k >= 1,
%F A184634 a(5*k+1) = 50*k^3 + 30*k^2 + 6*k for k >= 1,
%F A184634 a(5*k+2) = 50*k^3 + 60*k^2 + 24*k + 3,
%F A184634 a(5*k+3) = 50*k^3 + 90*k^2 + 54*k + 10 for k >= 1,
%F A184634 a(5*k+4) = 50*k^3 + 120*k^2 + 96*k + 25.
%F A184634 This implies the conjectured recurrence for n >= 12.  (End)
%p A184634 f:= proc(n) local k, t;
%p A184634   t:= n mod 5;
%p A184634   k:= (n-t)/5;
%p A184634   [50*k^3, 50*k^3 + 30*k^2 + 6*k,
%p A184634 50*k^3 + 60*k^2 + 24*k + 3,
%p A184634 50*k^3 + 90*k^2 + 54*k + 10,
%p A184634 50*k^3 + 120*k^2 + 96*k + 25][t+1]
%p A184634 end proc:
%p A184634 f(1):= 1: f(3):= 11:
%p A184634 map(f, [$1..100]); # _Robert Israel_, Feb 25 2019
%t A184634 p[n_]:=FractionalPart[(n^4+10)^(1/4)];
%t A184634 q[n_]:=Floor[1/p[n]];
%t A184634 Table[q[n], {n, 1, 80}]
%t A184634 FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
%t A184634 Join[{1,3,11},LinearRecurrence[{3,-3,1,0,1,-3,3,-1},{25,50,86,137,204,291,400,532},67]] (* _Ray Chandler_, Aug 02 2015 *)
%o A184634 (PARI) a(n)=1\frac(sqrtn(n^4+10,4)) \\ _Charles R Greathouse IV_, Feb 07 2016
%Y A184634 Cf. A184536.
%K A184634 nonn
%O A184634 1,2
%A A184634 _Clark Kimberling_, Jan 18 2011