|
%I #60 Dec 26 2018 03:38:09
%S 0,1,0,2,1,2,0,3,4,2,1,3,0,4,6,5,2,3,1,4,0,5,8,6,7,3,2,4,1,5,0,6,10,7,
%T 9,8,3,4,2,5,1,6,0,7,12,8,11,9,10,4,3,5,2,6,1,7,0,8,14,9,13,10,12,11,
%U 4,5,3,6,2,7,1,8,0,9,16,10,15,11,14,12,13,5,4,6,3,7,2,8,1,9,0,10,18,11,17,12,16,13,15,14
%N Triangle read by rows T(n,k) in which row n lists the first 2n-1 nonnegative integers representing 2n-1 equidistant points labeled in counterclockwise direction around a circle, with the 0th point at the top and reading them from left to right.
%H James Bentley, <a href="/A276669/b276669.txt">Rows 0...405 flattened</a>
%e Triangle begins:
%e 0;
%e 1, 0, 2;
%e 1, 2, 0, 3, 4;
%e 2, 1, 3, 0, 4, 6, 5;
%e 2, 3, 1, 4, 0, 5, 8, 6, 7;
%e 3, 2, 4, 1, 5, 0, 6, 10, 7, 9, 8;
%e 3, 4, 2, 5, 1, 6, 0, 7, 12, 8, 11, 9, 10;
%e 4, 3, 5, 2, 6, 1, 7, 0, 8, 14, 9, 13, 10, 12, 11;
%e 4, 5, 3, 6, 2, 7, 1, 8, 0, 9, 16, 10, 15, 11, 14, 12, 13;
%e 5, 4, 6, 3, 7, 2, 8, 1, 9, 0, 10, 18, 11, 17, 12, 16, 13, 15, 14;
%e ...
%e Illustration of numbers around a circle associated to the fourth row of triangle:
%e .
%e . 0
%e . 1 6
%e .
%e . 2 5
%e .
%e . 3 4
%e .
%e So the 4th row of the triangle is [2, 1, 3, 0, 4, 6, 5].
%t f[n_] := Transpose[ Sort[ Table[{Sin[ 2i*Pi/n + Pi], i}, {i, 0, n -1}], #1[[1]] < #2[[1]] &]][[2]]; Table[ f[n], {n, 1, 19, 2}] // Flatten (* _Robert G. Wilson v_, Nov 18 2016 *) (* Changing the constant in the Mmca coding changes where the reading begins. Pi starts it at the 9 o'clock position, Pi/2 would start it at the 12 o'clock position, 0 would have it start at the 3 o'clock position, etc. *)
%Y Column 1 is A110654. Middle diagonal is A000004. Apart from the first term, right border gives A001651.
%Y Cf. A001477, A004526.
%K nonn,tabf,look
%O 1,4
%A _James Bentley_ and _Omar E. Pol_, Sep 12 2016
|
