cp's OEIS Frontend

A321905 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 A321905 #16 Feb 09 2020 02:42:51
%S A321905 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,
%T A321905 3,13,31,1,59,53,55,9,51,61,47,17,43,5,39,25,35,13,31,33,27,21,23,41,
%U A321905 19,29,15,49,11,37,7,57,3,45,63
%N A321905 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 A321905 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 A321905 The n-th row contains 2^(n-1) numbers, and is a permutation of the odd numbers below 2^n.
%C A321905 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 A321905 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.
%C A321905 T(n,k) is the multiplicative inverse of A321904(n,k) modulo 2^n.
%F A321905 T(n,k) = 2^n - A321906(n,2^(n-1)-1-k).
%e A321905 Table starts
%e A321905 1,
%e A321905 1, 3,
%e A321905 1, 3, 5, 7,
%e A321905 1, 11, 5, 7, 9, 3, 13, 15,
%e A321905 1, 27, 21, 23, 9, 19, 29, 15, 17, 11, 5, 7, 25, 3, 13, 31,
%e A321905 1, 59, 53, 55, 9, 51, 61, 47, 17, 43, 5, 39, 25, 35, 13, 31, 33, 27, 21, 23, 41, 19, 29, 15, 49, 11, 37, 7, 57, 3, 45, 63,
%e A321905 ...
%o A321905 (PARI) T(n, k) = my(m=1); while(Mod(2*k+1, 2^n)^m!=m, m+=2); m
%o A321905 tabf(nn) = for(n=1, nn, for(k=0, 2^(n-1)-1, print1(T(n, k), ", ")); print)
%Y A321905 Cf. A007814.
%Y A321905 {x^x} and its inverse: A320561 & A320562.
%Y A321905 {x^(-x)} and its inverse: A321901 & A321904.
%Y A321905 {x^(1/x)} and its inverse: A321902 & this sequence.
%Y A321905 {x^(-1/x)} and its inverse: A321903 & A321906.
%K A321905 nonn,tabf
%O A321905 1,3
%A A321905 _Jianing Song_, Nov 21 2018