A321902 Irregular table read by rows: T(n,k) = (2*k+1)^(1/(2*k+1)) mod 2^n, 0 <= k <= 2^(n-1) - 1.

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

Offset

1,3

Comments

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

For n >= 3, T(n,k) = 2*k + 1 iff k is divisible by 2^floor((n-1)/2) or k = 2^(n-2) - 1 or k = 2^(n-1) - 1.

T(n,k) is the multiplicative inverse of A321903(n,k) modulo 2^n.

Links

Table of n, a(n) for n=1..63.

Formula

T(n,k) = 2^n - A321903(n,2^(n-1)-1-k).

Example

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

Prog

(PARI) T(n, k) = my(m=1); while(Mod(m, 2^n)^(2*k+1)!=2*k+1, m+=2); m
tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print)

Crossrefs

Cf. A007814.

{x^x} and its inverse: A320561 & A320562.

{x^(-x)} and its inverse: A321901 & A321904.

{x^(1/x)} and its inverse: this sequence & A321905.

{x^(-1/x)} and its inverse: A321903 & A321906.

Sequence in context: A189442 A320562 A320561 * A321905 A323495 A323554

Adjacent sequences: A321899 A321900 A321901 * A321903 A321904 A321905

Keyword

nonn,tabf

Author

Jianing Song, Nov 21 2018

Status

approved