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%I #33 Nov 07 2017 21:07:00
%S 1,4,23,26,45,48,67,70,89,92,111,114,133,136,155,158,177,180,183,199,
%T 202,205,221,224,227,243,246,249,265,268,271,290,293,312,315,334,337,
%U 356,359,378,381,400,403,422,425,444,447,466,469,488,491,510,513,532,535,538,554,557,560,576,579,582,598,601,604,620
%N Values of positive integer i such that floor(tan(i)) = 1.
%C The sequence is the first result in the chain of iteration leading to the ultimate sequence A258024.
%C Sequence terms are also the roots of A000503(i)=1, starting from i=1.
%C This is a subsequence of A258024 from which this differs for the first time at n=11, where a(11) = 111, while A258024(11) = 105, the term not included in this sequence. Note that A000503(105) = 4, a term which is included in this sequence. - _Antti Karttunen_, Oct 30 2017
%C Numbers k such that Pi/4 <= k - m*Pi < arctan(2) for some m. - _Robert Israel_, Nov 06 2017
%H Robert Israel, <a href="/A293698/b293698.txt">Table of n, a(n) for n = 1..10000</a>
%e The values of floor(tan(i)), starting from i=0, are given in A000503. Those i, for which floor(tan(i))=1 is true, are the roots of this equation. Thus the roots are the positions of 1 in A000503(i>0).
%e For n=1, i=1; a(1)=1.
%e For n=2, i=4; a(2)=4.
%e For n=3, i=23; a(3)=23.
%t rootsp = Flatten[Position[Table[Floor[Tan[i]], {i, 1, 10^6}], 1]
%t (*a(n) = rootsp[[n]]*)
%t Alternatively:
%t rootsp = {}; Do[If[Floor[Tan[n]] == 1, AppendTo[rootsp, n]], {n, 1, 10^6}]
%t rootsp (*a(n) = rootsp[[n]]*)
%t Select[ Range@ 622, Floor@ Tan@ # == 1 &] (* _Robert G. Wilson v_, Nov 06 2017 *)
%o (PARI) isok(n) = floor(tan(n)) == 1; \\ _Michel Marcus_, Oct 24 2017
%o (PARI) first(n) = {my(res = vector(n), i = 0, pi = [Pi, Pi], sols = [atan(1), atan(2)]); while(1, for(j = ceil(sols[1]), floor(sols[2]), i++; if(i>n, return(res)); res[i] = j); sols+=[Pi(), Pi()])} \\ _David A. Corneth_, Oct 24 2017
%Y Cf. A000503, A258024, A293751, A293700, A293701, A293704, A293699, A293702, A293705, A004112, A024814.
%K nonn
%O 1,2
%A _V.J. Pohjola_, Oct 15 2017
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