OFFSET
1,1
COMMENTS
For all n, there are at least 6 points where the transit of circumference occurs exactly at the corners. The rare case is when the transit occurs at 2 corners of a triangle, i.e., at n = 1, 13, 181, 35113, ... , (A001570(n)). The pattern repeats itself at every Pi/3 sector along the circumference. The triangle count per half sector by rows can be arranged as an irregular triangle as shown in the illustration. The rows count (A242396) is equal to the case centered at (1/2,0), A242395.
LINKS
Kival Ngaokrajang, Illustration of initial terms
Kival Ngaokrajang, Illustration for rare cases
PROG
(Small Basic)
For n =1 To 100
r6=n*math.Sin(30*Math.Pi/180)/(Math.Power(3, 0.5)/2)
r6a=math.Round(r6)
If r6-math.Floor(r6) >0.5 Then
last=1
Else
last=2
EndIf
'find corner intersecting points-----------------------
k=0
ic=0
h=Math.Power(1-0.5*0.5, 0.5)
c=math.Floor(n/h)
For i = h To c Step h
For j = 0.5 To n Step 0.5
r=Math.Power(i*i+j*j, 0.5)
If r = n Then
k=k+1
EndIf
EndFor
EndFor
if k > 1 then
ic=math.Floor(k/3)
EndIf
'------------------------------------------------------
a=0
b=0
For ii=1 To r6a
If ii=1 Then
a=a+1
Else
If ii = r6a Then
a=a+last
Else
a=a+2
EndIf
EndIf
b=a
EndFor
if n =1 then
aa = 1
Else
aa =1*(a-2*ic)*2+1
endif
TextWindow.Write(6*aa+", ")
EndFor
CROSSREFS
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, May 13 2014
STATUS
approved