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A216320 Irregular triangle: row n lists the Modd n order of the odd members of the reduced smallest nonnegative residue class modulo n. 5
1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 3, 3, 1, 4, 4, 2, 1, 3, 3, 1, 4, 4, 2, 1, 5, 5, 5, 5, 1, 2, 2, 2, 1, 3, 2, 6, 3, 6, 1, 3, 6, 3, 6, 2, 1, 4, 2, 4, 1, 8, 8, 4, 4, 8, 8, 2, 1, 8, 8, 8, 4, 8, 2, 4, 1, 6, 6, 3, 3, 2, 1, 9, 9, 3, 9, 3, 9, 9, 9, 1, 4, 4, 2, 2, 4, 4, 2, 1, 3, 6, 2, 3, 6 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

The length of row n is delta(n):=A055034(n).

For the multiplicative group Modd n see a comment on A203571, and also on A216319.

A216319(n,k)^a(n,k) == +1 (Modd n), n >= 1.

If the Modd n order of an (odd) element from row n of A216319 is delta(n) (the row length) then this element is a primitive root Modd n. There is no primitive root Modd n if no such element of order delta(n) exists. For example, n = 12, 20, ... (see A206552 for more of these n values). There are phi(delta(n)) = A216321(n) such primitive roots Modd n if there exists one, where phi=A000010 (Euler's totient). The multiplicative group Modd n is cyclic if and only if there exists a primitive root Modd n. The multiplicative group Modd n is isomorphic to the Galois group G(Q(rho(n)/Q) with the algebraic number rho(n) := 2*cos(Pi/n), n>=1.

LINKS

Table of n, a(n) for n=1..93.

FORMULA

a(n,k) = order of A216319(n,k) Modd n, n>=1, k=1, 2, ..., A055034(n). This means: A216319(n,k)^a(n,k) == +1 (Modd n), n>=1, and a(n,k) is the smallest positive integer exponent satisfying this congruence. For Modd n see a comment on A203571.

EXAMPLE

The table a(n,k) begins:

n\k 1  2  3  4  5  6  7  8  9 ...

1   1

2   1

3   1

4   1  2

5   1  2

6   1  2

7   1  3  3

8   1  4  4  2

9   1  3  3

10  1  4  4  2

11  1  5  5  5  5

12  1  2  2  2

13  1  3  2  6  3  6

14  1  3  6  3  6  2

15  1  4  2  4

16  1  8  8  4  4  8  8  2

17  1  8  8  8  4  8  2  4

18  1  6  6  3  3  2

19  1  9  9  3  9  3  9  9  9

20  1  4  4  2  2  4  4  2

...

a(7,2) = 3 because A216319(7,2) = 3 and 3^1 == 3 (Modd 7);

  3^2 = 9 == 5 (Modd 7) because floor(9/7)= 1 which is odd, therefore 9 (Modd 7) = -9 (mod 7) = 5; 3^3 == 5*3 (Modd n)

  = +1 because floor(15/7)=2 which is even, therefore 15 (Modd 7) = 15 (modd 7) = +1.

Row n=12 is the first row without an order = delta(n) (row length), in this case 4. Therefore there is no primitive root Modd 12, and the multiplicative group Modd 12 is non-cyclic.

  Its cycle structure is [[5,1],[7,1],[11,1]] which is the group Z_2 x Z_2 (the Klein 4-group).

CROSSREFS

Cf. A203571, A216321.

Sequence in context: A185215 A259551 A058232 * A308455 A214716 A116933

Adjacent sequences:  A216317 A216318 A216319 * A216321 A216322 A216323

KEYWORD

nonn,easy,tabf

AUTHOR

Wolfdieter Lang, Sep 21 2012

STATUS

approved

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Last modified August 6 10:14 EDT 2020. Contains 336245 sequences. (Running on oeis4.)