login
a(n) = floor( 1/(1 - cos(Pi/n)) ).
4

%I #16 Nov 11 2024 21:20:15

%S 0,1,2,3,5,7,10,13,16,20,24,29,34,39,45,52,58,65,73,81,89,98,107,116,

%T 126,137,147,159,170,182,194,207,220,234,248,262,277,292,308,324,340,

%U 357,374,392,410,428,447,467,486,506,527,548,569,591,613,635,658

%N a(n) = floor( 1/(1 - cos(Pi/n)) ).

%C This sequence provides insight into the manner of convergence of the sequence cos(Pi/n).

%H Clark Kimberling, <a href="/A248360/b248360.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) ~ 2*n^2/Pi^2. - _Vaclav Kotesovec_, Oct 09 2014

%e Approximations:

%e n ... 1-cos(Pi/n) ... 1/(1-cos(Pi/n))

%e 1 ... 2 ............. 0.5

%e 2 ... 1 ............. 1

%e 3 ... 0.5 ........... 2

%e 4 ... 0.292893 ...... 3.31421

%e 5 ... 0.190983 ...... 5.23607

%e 6 ... 0.133975 ...... 7.4741

%t z = 800; f[n_] := f[n] = Select[Range[z], Cos[Pi/#] + 1/(#*n) > 1 &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248359 *)

%t Table[Floor[1/(1 - Cos[Pi/n])], {n, 1, z/10}] (* A248360 *)

%Y Cf. A248360.

%K nonn,easy

%O 1,3

%A _Clark Kimberling_, Oct 07 2014