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A250211 Square array read by antidiagonals: A(m,n) = multiplicative order of m mod n, or 0 if m and n are not coprime. 3
1, 1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 0, 0, 0, 1, 1, 1, 1, 2, 4, 1, 1, 0, 2, 0, 4, 0, 1, 1, 1, 0, 1, 2, 0, 3, 1, 1, 0, 1, 0, 0, 0, 6, 0, 1, 1, 1, 2, 2, 1, 2, 3, 2, 6, 1, 1, 0, 0, 0, 4, 0, 6, 0, 0, 0, 1, 1, 1, 1, 1, 4, 1, 2, 2, 3, 4, 10, 1, 1, 0, 2, 0, 2, 0, 0, 0, 6, 0, 5, 0, 1, 1, 1, 0, 2, 0, 0, 1, 2, 0, 0, 5, 0, 12, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,9

COMMENTS

Read by antidiagonals:

m\n   1    2    3    4    5    6    7    8    9    10    11    12    13

1     1    1    1    1    1    1    1    1    1     1     1     1     1

2     1    0    2    0    4    0    3    0    6     0    10     0    12

3     1    1    0    2    4    0    6    2    0     4     5     0     3

4     1    0    1    0    2    0    3    0    3     0     5     0     6

5     1    1    2    1    0    2    6    2    6     0     5     2     4

6     1    0    0    0    1    0    2    0    0     0    10     0    12

7     1    1    1    2    4    1    0    2    3     4    10     2    12

8     1    0    2    0    4    0    1    0    2     0    10     0     4

9     1    1    0    1    2    0    3    1    0     2     5     0     3

10    1    0    1    0    0    0    6    0    1     0     2     0     6

11    1    1    2    2    1    2    3    2    6     1     0     2    12

12    1    0    0    0    4    0    6    0    0     0     1     0     2

13    1    1    1    1    4    1    2    2    3     4    10     1     0

etc.

A(m,n) = Least k>0 such that m^k=1 (mod n), or 0 if no such k exists.

It is easy to prove that column n has period n.

A(1,n) = 1, A(m,1) =1.

If A(m,n) differs from 0, it is period length of 1/n in base m.

The maximum number in column n is psi(n) (A002322(n)), and all numbers in column n (except 0) divide psi(n), and all factors of psi(n) are in column n.

Except the first row, every row contains all natural numbers.

LINKS

Table of n, a(n) for n=1..105.

EXAMPLE

A(3,7) = 6 because:

3^0 = 1 (mod 7)

3^1 = 3 (mod 7)

3^2 = 2 (mod 7)

3^3 = 6 (mod 7)

3^4 = 4 (mod 7)

3^5 = 5 (mod 7)

3^6 = 1 (mod 7)

...

And the period is 6, so A(3,7) = 6.

MAPLE

f:= proc(m, n)

  if igcd(m, n) <> 1 then 0

  elif n=1 then 1

  else numtheory:-order(m, n)

  fi

end proc:

seq(seq(f(t-j, j), j=1..t-1), t=2..65); # Robert Israel, Dec 30 2014

MATHEMATICA

a250211[m_, n_] = If[GCD[m, n] == 1, MultiplicativeOrder[m, n], 0]

Table[a250211[t-j, j], {t, 2, 65}, {j, 1, t-1}]

CROSSREFS

Cf. A002322, A111076, A111725, A001918, A008330, A007733, A002326, A007732, A051626, A066799.

See A139366 for another version.

Sequence in context: A080080 A093662 A284256 * A243753 A219238 A025918

Adjacent sequences:  A250208 A250209 A250210 * A250212 A250213 A250214

KEYWORD

nonn,easy,tabl

AUTHOR

Eric Chen, Dec 29 2014

STATUS

approved

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Last modified April 7 15:56 EDT 2020. Contains 333306 sequences. (Running on oeis4.)