login
Irregular table read by rows: T(n,k) is the smallest m such that m^(-1/m) == 2*k + 1 (mod 2^n), 0 <= k <= 2^(n-1) - 1.
7

%I #15 Feb 09 2020 02:42:54

%S 1,1,3,1,3,5,7,1,3,13,7,9,11,5,15,1,19,29,7,25,27,21,15,17,3,13,23,9,

%T 11,5,31,1,19,61,7,57,27,53,15,49,35,45,23,41,43,37,31,33,51,29,39,25,

%U 59,21,47,17,3,13,55,9,11,5,63

%N Irregular table read by rows: T(n,k) is the smallest m such that m^(-1/m) == 2*k + 1 (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 (2*k+1)^(-m) == m (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 For n >= 3, T(n,k) = 2*k + 1 iff k == -1 (mod 2^floor((n-1)/2)) or k = 0 or k = 2^(n-2).

%C T(n,k) is the multiplicative inverse of A320562(n,k) modulo 2^n.

%F T(n,k) = 2^n - A321905(n,2^(n-1)-1-k).

%e Table starts

%e 1,

%e 1, 3,

%e 1, 3, 5, 7,

%e 1, 3, 13, 7, 9, 11, 5, 15,

%e 1, 19, 29, 7, 25, 27, 21, 15, 17, 3, 13, 23, 9, 11, 5, 31,

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

%e ...

%o (PARI) T(n, k) = my(m=1); while(Mod(2*k+1, 2^n)^(-m)!=m, m+=2); m

%o tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print)

%Y Cf. A007814.

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

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

%Y {x^(1/x)} and its inverse: A321902 & A321905.

%Y {x^(-1/x)} and its inverse: A321903 & this sequence.

%K nonn,tabf

%O 1,3

%A _Jianing Song_, Nov 21 2018