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%I #12 Jul 26 2023 03:16:40
%S 1,1,2,1,3,3,1,4,2,4,1,5,5,5,5,1,6,3,2,3,6,1,7,7,7,7,7,7,1,8,4,8,2,8,
%T 4,8,1,9,9,3,9,9,3,9,9,1,10,5,10,5,2,5,10,5,10,1,11,11,11,11,11,11,11,
%U 11,11,11,1,12,6,4,3,12,2,12,3,4,6,12
%N Triangular array T read by rows: T(n,k) gives the additive orders k modulo n, for k = 0,1, ..., n-1.
%C As a sequence A054531(n) = a(n+1), n >= 1.
%C As a triangular array this is the row reversed version of A054531.
%C The additive order of an element x of a group (G, +) is the least positive integer j with j*x := x + x + ... + x (j summands) = 0.
%C Equals A106448 when the first column (k = 0) of ones is removed. - _Georg Fischer_, Jul 26 2023
%H Indranil Ghosh, <a href="/A277227/b277227.txt">Rows 1..100 of triangle, flattened</a>
%F T(n, k) = order of the elements k of the finite abelian group (Z/(n Z), +), for k = 0, 1, ..., n-1.
%F T(n, k) = n/GCD(n, k), n >= 1, k = 0, 1, ..., n-1.
%F T(n, k) = A054531(n, n-k), n >=1, k = 0, 1, ..., n-1.
%e The triangle begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 ...
%e 1: 1
%e 2: 1 2
%e 3: 1 3 3
%e 4: 1 4 2 4
%e 5: 1 5 5 5 5
%e 6: 1 6 3 2 3 6
%e 7: 1 7 7 7 7 7 7
%e 8: 1 8 4 8 2 8 4 8
%e 9: 1 9 9 3 9 9 3 9 9
%e 10: 1 10 5 10 5 2 5 10 5 10
%e 11: 1 11 11 11 11 11 11 11 11 11 11
%e 12: 1 12 6 4 3 12 2 12 3 4 6 12
%e ...
%e T(n, 0) = 1*0 = 0 = 0 (mod n), and n/GCD(n,0) = n/n = 1.
%e T(4, 2) = 2 because 2 + 2 = 4 = 0 (mod 4) and 2 is not 0 (mod 4).
%e T(4, 2) = n/GCD(2, 4) = 4/2 = 2.
%Y Cf. A054531, A106448.
%K nonn,tabl,easy
%O 1,3
%A _Wolfdieter Lang_, Oct 20 2016
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