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A319665
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Irregular triangle read by rows: T(n,k) = log_5(4*k + 1) mod 2^n, n >= 2, 0 <= k <= 2^(n-2) - 1.
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1
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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, 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, 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
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OFFSET
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2,6
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COMMENTS
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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.
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.
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.
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LINKS
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EXAMPLE
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Table begins
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,
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,
...
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PROG
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(PARI) T(n, k) = my(i=0); while(Mod(5, 2^n)^i!=4*k+1, i++); i
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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