A250211 Square array read by antidiagonals: A(m,n) = multiplicative order of m mod n, or 0 if m and n are not coprime.

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

Offset: 1

Author

Eric Chen, Dec 29 2014

Keywords

• nonn
• easy
• tabl

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.

Examples

			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.
		

Crossrefs

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

See A139366 for another version.

Programs

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}]