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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A337938 Irregular triangle read by rows: T(n, k) gives the primitive period of the sequence {k (Modd n)}_{k >= 0}, for n >= 1.

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%I A337938 #12 Apr 27 2021 11:51:28
%S A337938 0,0,1,0,1,2,0,2,1,0,1,2,3,0,3,2,1,0,1,2,3,4,0,4,3,2,1,0,1,2,3,4,5,0,
%T A337938 5,4,3,2,1,0,1,2,3,4,5,6,0,6,5,4,3,2,1,0,1,2,3,4,5,6,7,0,7,6,5,4,3,2,
%U A337938 1,0,1,2,3,4,5,6,7,8,0,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9,0,9,8,7,6,5,4,3,2,1
%N A337938 Irregular triangle read by rows: T(n, k) gives the primitive period of the sequence {k (Modd n)}_{k >= 0}, for n >= 1.
%C A337938 The length of row n is 1 for n = 1, 2 for n = 2, and 2*n for n >= 3.
%C A337938 The modified modular equivalence relation Modd n is defined, for integer k and positive integer n, by  k (Modd n) = k (mod n) if floor(k/n) is even, and -k (mod n) if floor(k/n) is odd. The smallest nonnegative complete residue system modulo n, namely RS(n) = {0, 1, ..., n-1}, is used. See the W. Lang link, Definition 4, eq. (69), p. 25 - 26.
%C A337938 In order to have row length 2*n for all n >= 1 one could use for n = 1 and 2 the imprimitive periods 0, 0 and 0, 1, 0, 1, respectively.
%C A337938 The name Modd n derives from the fact that the multiplicative (but not additive ) group Modd n has the smallest positive reduced residue system with only odd numbers, named RRSodd(n), as elements (for n = 0 RRS(n) = {0}, but here it is taken as {1}). This group is isomorphic to the Galois group G(rho(n)) = Gal(Q(rho(n))/Q), with rho(n) = 2*cos(pi/n). See the W. Lang link.
%H A337938 Wolfdieter Lang, <a href="http://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012, 2017.
%F A337938 T(n,k) = k (Modd n), for n >= 1, and k = 0 for n = 1, k = 0, 1 for n = 2, and k = 0, 1, ..., 2*n - 1, for n >= 3. For k (Modd n) see the comment above.
%e A337938 The irregular triangle begins:
%e A337938 n \ k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ..
%e A337938 1:    0
%e A337938 2:    0 1
%e A337938 3:    0 1 2 0 2 1
%e A337938 4:    0 1 2 3 0 3 2 1
%e A337938 5:    0 1 2 3 4 0 4 3 2 1
%e A337938 6:    0 1 2 3 4 5 0 5 4 3  2  1
%e A337938 7:    0 1 2 3 4 5 6 0 6 5  4  3  2  1
%e A337938 8:    0 1 2 3 4 5 6 7 0 7  6  5  4  3  2  1
%e A337938 9:    0 1 2 3 4 5 6 7 8 0  8  7  6  5  4  3  2  1
%e A337938 10   :0 1 2 3 4 5 6 7 8 9  0  9  8  7  6  5  4  3  2  1
%e A337938 ...
%e A337938 T(1, 0) = 0 because {k (Modd 1)}_{k >= 0} is the 0 sequence A000007:  0 (Modd 1) =  0 (mod 1) = 0, 1 (Modd 1) = -1 (mod 1) = 0,  2 (Modd 1) = 2 (mod 1) = 0, ... .
%e A337938 T(7, 6) = 6 because floor(6/7) = 0, which is even, hence 6 (Modd 7) = 6 (mod 7) = 6.
%e A337938 T(7, 8) = 6 because  floor(8/7) = 1, which is odd, hence  8 (Modd 7) = -8 (mod 7) = 6.
%Y A337938 Cf. Periodic sequences for n = 1, 2, ..., 7: A000007, A000035, A193680, A193682, A203571, A203572.
%Y A337938 Cf. A002262 (for mod n), A053616 (as a triangle, for mod* n).
%K A337938 nonn,tabf,easy
%O A337938 1,6
%A A337938 _Wolfdieter Lang_, Oct 25 2020