%I #54 Feb 22 2020 02:13:11
%S 2,3,5,8,10,11,13,16,18,21,23,24,26,29,31,32,34,36,37,39,42,44,45,47,
%T 50,52,53,55,57,58,60,63,65,66,68,71,73,76,78,79,81,84,86,87,89,91,92,
%U 94,97,99,100,102,105,107,110,112,113,115,118,120,121,123,126,128,131,133,134,136,139,141,142,144,146
%N Numbers k such that {k*phi} < 0.25 or {k*phi} > 0.75, where phi is the golden ratio (1 + sqrt(5))/2 and { } denotes fractional part.
%C Numbers k such that k rotations by the golden angle yields a result between -Pi/2 and Pi/2 radians.
%H Michael De Vlieger, <a href="/A295085/b295085.txt">Table of n, a(n) for n = 0..10000</a>
%H Burghard Herrmann, <a href="https://www.fq.math.ca/Papers1/57-5/herrmann.pdf">How integer sequences find their way into areas outside pure mathematics</a>, The Fibonacci Quarterly (2019) Vol. 57, No. 5, 67-71.
%H P. Prusinkiewicz and A. Lindenmayer, <a href="http://www.algorithmicbotany.org/papers/abop/abop-ch4.pdf">Chapter 4, Phyllotaxis</a>, The Algorithmic Beauty of Plants (1990).
%t Select[Range@ 150, Or[# < 1/4, # > 3/4] &@ FractionalPart[# GoldenRatio] &] (* _Michael De Vlieger_, Nov 15 2017 *)
%o (R) Phi=(sqrt(5)+1)/2 # Golden ratio
%o fp=function(x) x-floor(x) # fractional part
%o M=200
%o alpha=fp((1:M)*Phi) # angles in turn
%o PF=c(); PB=c() # Phyllotaxis front and back
%o for (i in 1:M) if ((alpha[i]>0.25)*(alpha[i]<0.75)) PB=c(PB,i) else PF=c(PF,i)
%o (PARI) isok(n) = my(phi=(1+sqrt(5))/2); (frac(n*phi)<1/4) || (frac(n*phi)>3/4); \\ _Michel Marcus_, Nov 14 2017
%Y Complement of A190250 (as has been proved), thus, intertwining of A190249 and A190251.
%Y Cf. A001622.
%K nonn,easy
%O 0,1
%A _Burghard Herrmann_, Nov 14 2017