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A235089
a(n)*Pi is the total length of irregular spiral (center points: 2, 1, 3, 4) after n rotations.
6
3, 10, 13, 20, 23, 30, 33, 40, 43, 50, 53, 60, 63, 70, 73, 80, 83, 90, 93, 100, 103, 110, 113, 120, 123, 130, 133, 140, 143, 150, 153, 160, 163, 170, 173, 180, 183, 190, 193, 200, 203, 210, 213, 220, 223, 230, 233, 240, 243, 250, 253, 260, 263, 270, 273, 280, 283, 290, 293, 300, 303, 310, 313, 320, 323, 330, 333
OFFSET
1,1
COMMENTS
Let points 2, 1, 3 & 4 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 and maintain continuity of circumferences. Continue using this procedure at point 3, 4, 1, ... and so on. The form is non-expanded loop.
The alternative point order [2, 3, 1, 4] gives the same pattern with reflection, but the sequence will be 2*A047215(n). See illustration in links.
Conjecture: Numbers equivalent 0 or 3 modulo 10. - Ralf Stephan, Jan 13 2014
FORMULA
Conjecture: a(n) = -1+(-1)^n+5*n. a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: x*(7*x+3) / ((x-1)^2*(x+1)). - Colin Barker, Jan 16 2014
PROG
(Small Basic)
a[1]=3
For n = 1 To 100
d1=3
If Math.Remainder(n+1, 2)=0 then
d1=7
EndIf
a[n+1]=a[n]+d1
TextWindow.Write(a[n]+", ")
EndFor
CROSSREFS
Cf. A014105*Pi (total spiral length, 2 inline center points). A234902*Pi, A234903*Pi, A234904*Pi (total spiral length, 3 inline center points).
Conjectured partial sums of A010705.
Sequence in context: A298418 A229205 A229483 * A022122 A042479 A042897
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, Jan 03 2014
STATUS
approved