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%I #7 Oct 01 2018 21:13:52
%S 0,0,1,0,1,2,3,0,1,6,7,4,5,2,3,0,1,6,15,12,13,2,11,8,9,14,7,4,5,10,3,
%T 0,1,6,15,28,13,2,27,24,25,30,7,20,5,26,19,16,17,22,31,12,29,18,11,8,
%U 9,14,23,4,21,10,3,0,1,6,47,28,45,2,59,56,25,62,7,20,5,58,19,48,49,54,31,12,29,50,43,40,9,46,55,4,53,42,3
%N Irregular triangle read by rows: T(n,k) = log_5(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 5^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, 1, 2, 3,
%e 0, 1, 6, 7, 4, 5, 2, 3,
%e 0, 1, 6, 15, 12, 13, 2, 11, 8, 9, 14, 7, 4, 5, 10, 3,
%e 0, 1, 6, 15, 28, 13, 2, 27, 24, 25, 30, 7, 20, 5, 26, 19, 16, 17, 22, 31, 12, 29, 18, 11, 8, 9, 14, 23, 4, 21, 10, 3,
%e ...
%o (PARI) T(n,k) = my(i=0); while(Mod(5,2^n)^i!=4*k+1, i++); i
%Y Cf. A007814, A319663.
%K nonn,tabf
%O 2,6
%A _Jianing Song_, Sep 25 2018
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