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Arrange numbers in a clockwise spiral with initial terms a(1)=1, a(2)=2, a(4)=3, a(6)=4, a(8)=5; thereafter each number shares a factor with each of its four (N,S,E,W) neighbors.
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%I #21 Aug 12 2018 21:27:53

%S 1,2,6,3,12,4,10,5,20,8,14,16,18,9,15,21,7,28,22,24,26,30,25,35,40,32,

%T 34,36,38,42,27,33,39,45,48,56,44,70,46,50,52,13,65,60,51,75,55,66,54,

%U 57,72,58,62,64,68,78,63,69,81,84,80,74,76,82,41

%N Arrange numbers in a clockwise spiral with initial terms a(1)=1, a(2)=2, a(4)=3, a(6)=4, a(8)=5; thereafter each number shares a factor with each of its four (N,S,E,W) neighbors.

%C Start with smallest number which has not yet appeared and satisfies the conditions: a(3)=6; thereafter always choose smallest number which has not yet appeared and satisfies the conditions.

%C This is a two-dimensional spiral analog of EKG sequence A064413.

%C In A064413 we have initial terms in the positions 1,2.

%C In the two-dimensional case we have 4 sides.

%C So the initial TERMS are

%C 5

%C 4 1 2 (1)

%C 3

%C But the POSITIONS in the spiral are indexed thus:

%C .

%C 7--8--9--10

%C |

%C 6 1--2

%C | |

%C 5--4--3

%C .

%C So the initial terms, by (1), are a(1)=1, a(2)=2, a(4)=3, a(6)=4, a(8)=5.

%C Conjecture: The sequence is a permutation of the positive integers. - _Vladimir Shevelev_, May 06 2015

%H Peter J. C. Moses, <a href="/A253279/b253279.txt">Table of n, a(n) for n = 1..5625</a>

%H Peter J. C. Moses, <a href="/A253279/a253279.pdf">The first few squares.</a>

%e The spiral begins

%e .

%e 26--30--25--35--40--32 etc.

%e |

%e 24 10---5--20---8

%e | | |

%e 22 4 1---2 14

%e | | | |

%e 28 12---3---6 16

%e | |

%e 7--21--15---9--18

%Y Cf. A064413, A257321-A257340, A257112.

%K nonn

%O 1,2

%A _Vladimir Shevelev_, May 02 2015

%E Correction of a(42) and more terms from _Peter J. C. Moses_, May 03 2015