%I #13 Jul 12 2014 16:46:03

%S 2,8,19,30,32,38,49,60,62,68,79,90,92,98,109,120,122,128,139,150,152,

%T 158,169,180,182,188,199,210,212,218,229,240,242,248,259,270,272,278,

%U 289,300,302,308,319,330,332,338,349,360,362,368,379,390,392,398,409,420,422,428,439,450,452,458,469,480,482,488

%N a(n)*Pi is the total length of irregular spiral (center points: 1, 2, 3, 4, 5; pattern 2) after n rotations.

%C Let points 1, 2, 3, 4 & 5 be placed on a straight line at intervals of 1 unit. At point 1 make a half unit circle then at point 2 make another half circle; by selecting radius point on the right hand side of point 1 (pattern 2); at point 3 make another half circle and maintain continuity of circumferences. Continue using this procedure at point 4, 5, 1, ... and so on.

%C Conjecture: All forms of 120 permutations 5 center points are non-expanded loops.

%H Kival Ngaokrajang, <a href="/A236327/a236327.pdf">Illustration of irregular spiral (center points: 1, 2, 3, 4, 5)</a>Pattern 2.

%F Conjecture from _Colin Barker_, Jul 12 2014: (Start)

%F a(n) = a(n-1)+a(n-4)-a(n-5).

%F G.f.: x*(11*x^3+11*x^2+6*x+2) / ((x-1)^2*(x+1)*(x^2+1)). (End)

%o (Small Basic)

%o n =5 'center points number 1<=n<=9

%o pt=1 'pattern1: pt=-1; pattern2: pt=1

%o i=12345 'center points order

%o rota=100 'rotations

%o sum=0

%o rc=1

%o r[1]=1

%o For i1 = 1 To n

%o d1=i/Math.Power(10,1)

%o i=math.Floor(d1)

%o d[i1]=(d1-i)*Math.Power(10,1)

%o EndFor

%o For j1=1 To n

%o For j2=1 To n

%o If d[j1]=j2 Then

%o dd[j2]=j1

%o endif

%o EndFor

%o EndFor

%o For j3=1 To n

%o If j3=n Then

%o dxy[j3]=dd[j3]-dd[1]

%o Else

%o dxy[j3]=dd[j3]-dd[j3+1]

%o EndIf

%o EndFor

%o For k1=1 To rota*n

%o cc=Math.Floor((k1-1)/n)

%o p[k1]=r[k1]+pt*dxy[k1-cc*n]*Math.Power(-1,Math.Remainder(k1,2))

%o r[k1+1]=p[k1]

%o sum=sum+math.Abs(r[k1])

%o If math.Abs(r[k1])>0 Then

%o rc=rc+1

%o EndIf

%o If rc=3 Then

%o TextWindow.Write(sum+", ")

%o rc=1

%o EndIf

%o EndFor

%Y Cf. A014105 (2 center points); A234902, A234903, A234904 (3 center points); A235088, A235089 (4 center points).

%K nonn

%O 1,1

%A _Kival Ngaokrajang_, Jan 22 2014