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Following the successive antidiagonals in A065188, let the n-th queen appear in square (x(n),y(n)); sequence gives x(n).
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%I #14 Aug 25 2016 23:06:23

%S 1,2,4,3,5,6,10,7,11,13,8,9,15,12,20,21,14,16,26,17,27,29,18,19,31,34,

%T 22,23,38,24,40,25,43,42,28,30,49,50,32,33,54,56,35,36,59,58,37,39,64,

%U 41,67,69,44,71,45,46,75,47,77,48,78,80,51,52,85,53,86,55,90,91,57,95,60,61,99,62,101,63

%N Following the successive antidiagonals in A065188, let the n-th queen appear in square (x(n),y(n)); sequence gives x(n).

%C See A275900 for y(n).

%C This is a permutation of the natural numbers.

%C This assumes the indexing starts at 1. See A275901, A275902 if the indexing begins at 0.

%H N. J. A. Sloane, <a href="/A275899/b275899.txt">Table of n, a(n) for n = 1..7500</a>

%p # To get the coordinates of queens in order of appearance; b8[] has list of terms of A065188

%p M:=7500; c1:=[]; c2:=[];

%p t1:=[seq(n+b8[n],n=1..M)];

%p t2:=sort(t1);

%p for n from 1 to M do

%p i:=t2[n]; member(i,t1,'j');

%p c1:=[op(c1),j]; c2:=[op(c2),b8[j]];

%p od:

%p c3:=map(x->x-1,c1):

%p c4:=map(x->x-1,c2):

%p [seq(c1[n],n=1..80)]; # A275899

%p [seq(c2[n],n=1..80)]; # A275900

%p [seq(c3[n],n=1..80)]; @ A275901

%p [seq(c4[n],n=1..80)]; @ A275902

%Y Cf. A065188, A275900, A275901, A275902.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Aug 24 2016