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

Author

Jianing Song, Nov 21 2018

Keywords

• nonn
• tabf

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.

Examples

			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,
...
		

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.

Programs

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)

Formula

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