

A323553


Irregular table read by rows: T(n,k) = (2*k+1)^(1/3) mod 2^n, 0 <= k <= 2^(n1)  1.


5



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, 11, 13, 31, 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, 1, 51, 117, 39, 41, 91, 29, 79, 81, 3, 69
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OFFSET

1,3


COMMENTS

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).
The nth row contains 2^(n1) numbers, and is a permutation of the odd numbers below 2^n.
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 2adic valuation of k.
T(n,k) is the multiplicative inverse of A323556(n,k) modulo 2^n.


LINKS

Table of n, a(n) for n=1..74.


EXAMPLE

Table starts
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, 5,23,25,11,13,31;
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;
..


MAPLE

A323553 := proc(n, k)
local x;
for x from 1 to 2^n1 by 2 do
if modp(x^3*(2*k+1), 2^n) = 1 then
return x;
end if;
end do:
end proc:
seq(seq(A323553(n, k), k=0..2^(n1)1), n=1..8) ; # R. J. Mathar, Dec 15 2020


CROSSREFS

Cf. A007814.
{(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).
Sequence in context: A256263 A006257 A323555 * A170898 A321901 A321906
Adjacent sequences: A323550 A323551 A323552 * A323554 A323555 A323556


KEYWORD

nonn,tabf


AUTHOR

Jianing Song, Aug 30 2019


EXTENSIONS

Values corrected following a suggestion from Don Reble  R. J. Mathar, Dec 15 2020


STATUS

approved



