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A320562
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Irregular table read by rows: T(n,k) is the smallest m such that m^m == 2*k + 1 (mod 2^n), 0 <= k <= 2^(n-1) - 1.
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8
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1, 1, 3, 1, 3, 5, 7, 1, 11, 5, 7, 9, 3, 13, 15, 1, 27, 21, 23, 9, 19, 29, 15, 17, 11, 5, 7, 25, 3, 13, 31, 1, 27, 21, 55, 9, 19, 29, 47, 17, 11, 37, 39, 25, 3, 45, 31, 33, 59, 53, 23, 41, 51, 61, 15, 49, 43, 5, 7, 57, 35, 13, 63
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OFFSET
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1,3
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COMMENTS
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The sequence {k^k mod 2^n} has period 2^n. The n-th row contains 2^(n-1) numbers, and is a permutation of the odd numbers below 2^n.
Note that the first 5 rows are the same as those in A320561, but after that they differ.
For all n, k we have v(T(n,k)-1, 2) = v(k, 2) + 1 and v(T(n,k)+1, 2) = v(k+1, 2) + 1, where v(k, 2) = A007814(k) is the 2-adic valuation of k. [Revised by Jianing Song, Nov 24 2018]
For n >= 3, T(n,k) = 2*k + 1 iff k is divisible by 2^floor((n-1)/2) or k = 2^(n-2) - 1 or k = 2^(n-1) - 1.
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LINKS
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FORMULA
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For given n >= 2 and 0 <= k <= 2^(n-2) - 1, T(n,k) = T(n-1,k) if T(n-1,k)^T(n-1,k) == 2*k + 1 (mod 2^n), otherwise T(n-1,k) + 2^(n-1); for 2^(n-2) <= k <= 2^(n-1) - 1, T(n,k) = T(n,k-2^(n-2)) + 2^(n-1) if T(n,k) < 2^(n-1), otherwise T(n,k-2^(n-2)) - 2^(n-1).
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EXAMPLE
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Table starts
1,
1, 3,
1, 3, 5, 7,
1, 11, 5, 7, 9, 3, 13, 15,
1, 27, 21, 23, 9, 19, 29, 15, 17, 11, 5, 7, 25, 3, 13, 31,
1, 27, 21, 55, 9, 19, 29, 47, 17, 11, 37, 39, 25, 3, 45, 31, 33, 59, 53, 23, 41, 51, 61, 15, 49, 43, 5, 7, 57, 35, 13, 63,
...
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MATHEMATICA
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Table[Block[{m = 1}, While[PowerMod[m, m, 2^n] != Mod[2 k + 1, 2^n], m++]; m], {n, 6}, {k, 0, 2^(n - 1) - 1}] // Flatten (* Michael De Vlieger, Oct 22 2018 *)
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PROG
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(PARI) T(n, k) = my(m=1); while(Mod(m, 2^n)^m!=2*k+1, m+=2); m
tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print);
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CROSSREFS
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{x^x} and its inverse: A320561 & this sequence.
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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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