A323553 Irregular table read by rows: T(n,k) = (2*k+1)^(-1/3) mod 2^n, 0 <= k <= 2^(n-1) - 1.

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

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 n-th row contains 2^(n-1) 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 2-adic 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^n-1 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^(n-1)-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