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Irregular table read by rows: T(n,k) = (2*k+1)^(1/5) mod 2^n, 0 <= k <= 2^(n-1) - 1.
4

%I #9 Aug 30 2019 21:50:18

%S 1,1,3,1,3,5,7,1,3,5,7,9,11,13,15,1,19,21,7,9,27,29,15,17,3,5,23,25,

%T 11,13,31,1,19,21,39,41,59,61,15,17,35,37,55,57,11,13,31,33,51,53,7,9,

%U 27,29,47,49,3,5,23,25,43,45,63,1,83,21,103,105,59,125,79,81,35,101,55,57,11,77,31,33,115,53,7,9,91,29,111,113,67,5,87,89,43,109,63

%N Irregular table read by rows: T(n,k) = (2*k+1)^(1/5) mod 2^n, 0 <= k <= 2^(n-1) - 1.

%C T(n,k) is the unique x in {1, 3, 5, ..., 2^n - 1} such that x^5 == 2*k + 1 (mod 2^n).

%C The n-th row contains 2^(n-1) numbers, and is a permutation of the odd numbers below 2^n.

%C 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.

%C T(n,k) is the multiplicative inverse of A323554(n,k) modulo 2^n.

%e Table starts

%e 1,

%e 1, 3,

%e 1, 3, 5, 7,

%e 1, 3, 5, 7, 9, 11, 13, 15,

%e 1, 19, 21, 7, 9, 27, 29, 15, 17, 3, 5, 23, 25, 11, 13, 31,

%e 1, 19, 21, 39, 41, 59, 61, 15, 17, 35, 37, 55, 57, 11, 13, 31, 33, 51, 53, 7, 9, 27, 29, 47, 49, 3, 5, 23, 25, 43, 45, 63,

%e ...

%o (PARI) T(n, k) = if(n==2, 2*k+1, lift(sqrtn(2*k+1+O(2^n),5)))

%o tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print)

%Y Cf. A007814.

%Y {(2*k+1)^e mod 2^n}: A323495 (e=-1), A323553 (e=-1/3), A323554 (e=-1/5), this sequence (e=1/5), A323556 (e=1/3).

%K nonn,tabf

%O 1,3

%A _Jianing Song_, Aug 30 2019