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Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.
1

%I #14 Mar 21 2017 10:46:52

%S 1,7,22,51,100,173,274,409,583,800,1064,1382,1757,2195,2700,3276,3930,

%T 4665,5487,6400,7408,8518,9733,11059,12500,14060,15746,17561,19511,

%U 21600,23832,26214,28749,31443,34300,37324,40522,43897,47455,51200,55136,59270,63605,68147,72900,77868,83058,88473,94119,100000

%N Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

%H Ray Chandler, <a href="/A184629/b184629.txt">Table of n, a(n) for n = 1..10000</a>

%H <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 a(n)=floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

%F 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) for n>=15.

%t p[n_]:=FractionalPart[(n^4+5)^(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, 7, 22, 51, 100, 173}, LinearRecurrence[{3, -3, 1, 0, 1, -3, 3, -1}, {274, 409, 583, 800, 1064, 1382, 1757, 2195}, 44]] (* _Ray Chandler_, Aug 01 2015 *)

%Y Cf. A184536.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jan 18 2011