cp's OEIS Frontend

A321906 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.

%I A321906 #15 Feb 09 2020 02:42:54
%S A321906 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 A321906 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 A321906 59,21,47,17,3,13,55,9,11,5,63
%N A321906 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 A321906 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 A321906 The n-th row contains 2^(n-1) numbers, and is a permutation of the odd numbers below 2^n.
%C A321906 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 A321906 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 A321906 T(n,k) is the multiplicative inverse of A320562(n,k) modulo 2^n.
%F A321906 T(n,k) = 2^n - A321905(n,2^(n-1)-1-k).
%e A321906 Table starts
%e A321906 1,
%e A321906 1, 3,
%e A321906 1, 3, 5, 7,
%e A321906 1, 3, 13, 7, 9, 11, 5, 15,
%e A321906 1, 19, 29, 7, 25, 27, 21, 15, 17, 3, 13, 23, 9, 11, 5, 31,
%e A321906 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 A321906 ...
%o A321906 (PARI) T(n, k) = my(m=1); while(Mod(2*k+1, 2^n)^(-m)!=m, m+=2); m
%o A321906 tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print)
%Y A321906 Cf. A007814.
%Y A321906 {x^x} and its inverse: A320561 & A320562.
%Y A321906 {x^(-x)} and its inverse: A321901 & A321904.
%Y A321906 {x^(1/x)} and its inverse: A321902 & A321905.
%Y A321906 {x^(-1/x)} and its inverse: A321903 & this sequence.
%K A321906 nonn,tabf
%O A321906 1,3
%A A321906 _Jianing Song_, Nov 21 2018