OFFSET
1,3
COMMENTS
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.
T(n,k) is the multiplicative inverse of A321906(n,k) modulo 2^n. - Jianing Song, Nov 24 2018
LINKS
Jianing Song, Table of n, a(n) for n = 1..8191 (Rows n=1..13)
FORMULA
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).
T(n,k) = 2^n - A321904(n,2^(n-1)-1-k). - Jianing Song, Nov 24 2018
EXAMPLE
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,
...
MATHEMATICA
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 *)
PROG
(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);
KEYWORD
nonn,tabf
AUTHOR
Jianing Song, Oct 15 2018
STATUS
approved
