%I #13 Sep 17 2024 19:07:16

%S 1,70,569,58704,15770314

%N Numbers k such that ceiling(2*Pi*k/sqrt(2)) != ceiling(Pi/arcsin(sqrt(2)/(2*k))).

%C 2*n/sqrt(2) > 1/arcsin(sqrt(2)/(2*n)) for all n > 0.

%C Limit_{x->oo} 2*x/sqrt(2) - 1/arcsin(sqrt(2)/(2*x)) = 0.

%e n k=a(n) 2*Pi*k/sqrt(2) Pi/arcsin(sqrt(2)/(2*k))

%e 1 1 4.44288293816 4.000000000000

%e 2 70 311.00180567109 310.996516371805

%e 3 569 2528.00039181211 2527.999741125982

%e 4 58704 260815.00000164873 260814.999995341832

%e 5 15770314 70065659.00000001744 70065658.999999993965

%Y Cf. A063448, A376066.

%Y Cf. A120702.

%K nonn,more

%O 1,2

%A _Hugo Pfoertner_, Sep 15 2024