%I #69 Jun 29 2022 11:44:39
%S 0,0,0,2,2,2,4,6,8,10,12,14,18,20,22,26,30,32,36,42,42,48,54,56,62,70,
%T 72,78,86,90,96,106,110,114,126,132,136,148,156,160,170,180,184,196,
%U 208,212,222,236,240,252,266,272,282,298,306,314,332,342,348,366,380
%N The number of squares, each vertex of which lies on the sinusoid y = n * sin(x), and whose centers coincide with the origin.
%C Graphically, the vertices of all squares are defined as the intersection points of the sinusoid y = n * sin(x) and the same sinusoid rotated 90 degrees around the origin.
%H Talmon Silver, <a href="/A345256/b345256.txt">Table of n, a(n) for n = 1..100</a>
%H Nicolay Avilov, <a href="https://www.diofant.ru/problem/3967/">Problem 2185. Squares and sinusoid</a> (in Russian).
%H Nicolay Avilov, <a href="https://elementy.ru/problems/2602/Kvadrat_v_sinuse">Problem Kvadrat v sinuse</a> (in Russian).
%H Talmon Silver, <a href="/A345256/a345256.txt">MUMPS-program</a>
%F a(n) = (K-1)/4, where K is the number of roots of n*sin(n*sin(x)) + x = 0.
%F From _Merab Leviashvili_, Jul 22 2021: (Start)
%F Writing d=floor(n/(Pi/2)) (mod 4), one has a(n)=
%F (floor(n/Pi))^2+2*floor(n*sin(n)/(2*Pi)+1/4) if d=0;
%F (floor(n/Pi+1/2))^2-1-2*floor(n*sin(n)/(2*Pi)+1/4) if d=1;
%F (floor(n/Pi))^2-1+2*floor(|n*sin(n)/(2*Pi)|+3/4) if d=2;
%F (floor(n/Pi+1/2))^2-2*floor(|n*sin(n)/(2*Pi)|+3/4) if d=3. (End)
%e a(3) = 0 since there are no squares with all their vertices on the curve y = 3*sin(x).
%e a(4) = 2 since there are 2 squares whose vertices lie on the curve y = 4*sin(x). The two squares have approximate coordinates: (1.02; 3.39), (3.39; -1.02), (-1.02; -3.39), (-3.39; 1.02) and (1.98; 3.67), (3.67; -1.98), (-1.98; -3.67), (-3.67; 1.98).
%K nonn
%O 1,4
%A _Nicolay Avilov_, Jun 12 2021
%E More terms from _Jinyuan Wang_, Jun 17 2021