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Irregular table read by rows: T(n,k) = (2*k+1)^(1/3) mod 2^n, 0 <= k <= 2^(n-1) - 1.
4

%I #9 Aug 30 2019 21:50:25

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

%T 3,5,31,1,59,29,23,25,19,53,47,49,43,13,7,9,3,37,31,33,27,61,55,57,51,

%U 21,15,17,11,45,39,41,35,5,63,1,123,93,23,25,83,53,47,49,43,13,71,73,3,101,95,97,91,61,119,121,51,21,15,17,11,109,39,41,99,69,63

%N Irregular table read by rows: T(n,k) = (2*k+1)^(1/3) 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^3 == 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 A323553(n,k) modulo 2^n.

%e Table starts

%e 1,

%e 1, 3,

%e 1, 3, 5, 7,

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

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

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

%e ...

%o (PARI) T(n, k) = if(n==2, 2*k+1, lift(sqrtn(2*k+1+O(2^n),3)))

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

%Y Cf. A322701, A322926, A322934, A322999.

%Y Cf. also A007814.

%Y {(2*k+1)^e mod 2^n}: A323495 (e=-1), A323553 (e=-1/3), A323554 (e=-1/5), A323555 (e=1/5), this sequence (e=1/3).

%K nonn,tabf

%O 1,3

%A _Jianing Song_, Aug 30 2019