%I #8 Oct 01 2018 21:14:05
%S 0,0,1,0,3,2,1,0,3,2,5,4,7,6,1,0,11,2,5,4,15,6,9,8,3,10,13,12,7,14,1,
%T 0,11,2,5,20,15,22,9,8,19,10,13,28,23,30,17,16,27,18,21,4,31,6,25,24,
%U 3,26,29,12,7,14,1,0,11,2,5,52,47,22,9,40,19,42,13,28,55,62,17,16,27,18,21,4,63,38,25,56,35,58,29,44,7,14,33
%N Irregular triangle read by rows: T(n,k) = log_(-3)(4*k + 1) mod 2^n, n >= 2, 0 <= k <= 2^(n-2) - 1.
%C The n-th row contains 2^(n-2) numbers. T(n,k) is the smallest e such that (-3)^e == 4*k + 1 (mod 2^n). This equation always has a solution in [0, 2^(n-2) - 1], so the n-th row is a permutation of 0, 1, 2, ..., 2^(n-2) - 1.
%C For e >= 4, the multiplicative order of a modulo 2^e equals to 2^(e-2) iff a == 3, 5 (mod 8); for e >= 5, the multiplicative order of a modulo 2^e equals to 2^(e-3) iff a == 7, 9 (mod 16); for e >= 6, the multiplicative order of a modulo 2^e equals to 2^(e-4) iff a == 15, 17 (mod 32), etc. From this we can see v(T(n,k), 2) = v(k, 2), where v(k, 2) = A007814(k) is the 2-adic valuation of k. Also, 4*k + 1 is a 2^v(k, 2)-th power residue but not a 2^(v(k, 2)+1)-th power residue modulo 2^i, i >= v(k, 2) + 3.
%C Define Chi(n,k) as: Chi(n,2*k) = 0 for all integers k, Chi(n,4*k+1) = exp(T(n,k)*Pi*i/2^(n-3)) for 0 <= k <= 2^(n-2) - 1 (i denotes the imaginary unit), Chi(n,4*k+3) = Chi(n,2^n-4*k-3) for 0 <= k <= 2^(n-2) - 1, Chi(n,2^n+k) = Chi(n,k) for all integers k, then Chi(n,k) forms a Dirichlet character modulo 2^n.
%e Table begins
%e 0,
%e 0, 1,
%e 0, 3, 2, 1,
%e 0, 3, 2, 5, 4, 7, 6, 1,
%e 0, 11, 2, 5, 4, 15, 6, 9, 8, 3, 10, 13, 12, 7, 14, 1,
%e 0, 11, 2, 5, 20, 15, 22, 9, 8, 19, 10, 13, 28, 23, 30, 17, 16, 27, 18, 21, 4, 31, 6, 25, 24, 3, 26, 29, 12, 7, 14, 1,
%e ...
%o (PARI) T(n,k) = my(i=0); while(Mod(-3,2^n)^i!=4*k+1, i++); i
%Y Cf. A007814, A319664.
%K nonn,tabf
%O 2,5
%A _Jianing Song_, Sep 25 2018