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%I #65 Jan 18 2015 06:18:04
%S 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,
%T 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,
%U 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
%N Square array read by antidiagonals: A(m,n) = multiplicative order of m mod n, or 0 if m and n are not coprime.
%C Read by antidiagonals:
%C m\n 1 2 3 4 5 6 7 8 9 10 11 12 13
%C 1 1 1 1 1 1 1 1 1 1 1 1 1 1
%C 2 1 0 2 0 4 0 3 0 6 0 10 0 12
%C 3 1 1 0 2 4 0 6 2 0 4 5 0 3
%C 4 1 0 1 0 2 0 3 0 3 0 5 0 6
%C 5 1 1 2 1 0 2 6 2 6 0 5 2 4
%C 6 1 0 0 0 1 0 2 0 0 0 10 0 12
%C 7 1 1 1 2 4 1 0 2 3 4 10 2 12
%C 8 1 0 2 0 4 0 1 0 2 0 10 0 4
%C 9 1 1 0 1 2 0 3 1 0 2 5 0 3
%C 10 1 0 1 0 0 0 6 0 1 0 2 0 6
%C 11 1 1 2 2 1 2 3 2 6 1 0 2 12
%C 12 1 0 0 0 4 0 6 0 0 0 1 0 2
%C 13 1 1 1 1 4 1 2 2 3 4 10 1 0
%C etc.
%C A(m,n) = Least k>0 such that m^k=1 (mod n), or 0 if no such k exists.
%C It is easy to prove that column n has period n.
%C A(1,n) = 1, A(m,1) =1.
%C If A(m,n) differs from 0, it is period length of 1/n in base m.
%C 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.
%C Except the first row, every row contains all natural numbers.
%e A(3,7) = 6 because:
%e 3^0 = 1 (mod 7)
%e 3^1 = 3 (mod 7)
%e 3^2 = 2 (mod 7)
%e 3^3 = 6 (mod 7)
%e 3^4 = 4 (mod 7)
%e 3^5 = 5 (mod 7)
%e 3^6 = 1 (mod 7)
%e ...
%e And the period is 6, so A(3,7) = 6.
%p f:= proc(m,n)
%p if igcd(m,n) <> 1 then 0
%p elif n=1 then 1
%p else numtheory:-order(m,n)
%p fi
%p end proc:
%p seq(seq(f(t-j,j),j=1..t-1),t=2..65); # _Robert Israel_, Dec 30 2014
%t a250211[m_, n_] = If[GCD[m, n] == 1, MultiplicativeOrder[m, n], 0]
%t Table[a250211[t-j, j], {t, 2, 65}, {j, 1, t-1}]
%Y Cf. A002322, A111076, A111725, A001918, A008330, A007733, A002326, A007732, A051626, A066799.
%Y See A139366 for another version.
%K nonn,easy,tabl
%O 1,9
%A _Eric Chen_, Dec 29 2014
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