A323553 - OEIS

%I #39 Jan 26 2024 11:30:26

%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 11,13,31,1,51,53,39,41,27,29,15,17,3,5,55,57,43,45,31,33,19,21,7,9,

%U 59,61,47,49,35,37,23,25,11,13,63,1,51,117,39,41,91,29,79,81,3,69

%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) == 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 A323556(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  1,51,53,39,41,27,29,15,17, 3, 5,55,57,43,45,31,33,19,21, 7,9,59,61,47,49,35,37,23,25,11,13,63;

%e  ...

%p A323553 := proc(n,k)

%p     local x;

%p     for x from 1 to 2^n-1 by 2 do

%p         if modp(x^3*(2*k+1),2^n) = 1 then

%p             return x;

%p         end if;

%p     end do:

%p end proc:

%p seq(seq(A323553(n,k),k=0..2^(n-1)-1),n=1..8) ; # _R. J. Mathar_, Dec 15 2020

%Y Cf. A007814.

%Y {(2*k+1)^e mod 2^n}: A323495 (e=-1), this sequence (e=-1/3), A323554 (e=-1/5), A323555 (e=1/5), A323556 (e=1/3).

%K nonn,tabf

%O 1,3

%A _Jianing Song_, Aug 30 2019

%E Values corrected following a suggestion from _Don Reble_ - _R. J. Mathar_, Dec 15 2020