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A345401 a(n) is the unique odd number h such that BCR(h*2^m-1) = 2n (except for BCR(0) = 1) where BCR is bit complement and reverse per A036044.

%I A345401 #58 Jul 20 2021 03:59:42
%S A345401 1,3,7,5,15,11,13,9,31,23,27,19,29,21,25,17,63,47,55,39,59,43,51,35,
%T A345401 61,45,53,37,57,41,49,33,127,95,111,79,119,87,103,71,123,91,107,75,
%U A345401 115,83,99,67,125,93,109,77,117,85,101,69,121,89,105,73,113,81,97,65,255,191
%N A345401 a(n) is the unique odd number h such that BCR(h*2^m-1) = 2n (except for BCR(0) = 1) where BCR is bit complement and reverse per A036044.
%C A345401 This sequence is a permutation of the odd numbers.
%C A345401 We have BCR(a(n)*2^m-1) = 2n when n = 0 for m >= 1, and BCR(a(n)*2^m-1) = 2n when n >= 1 for m >= 0.
%C A345401 Why this exception when n = 0? As a(0) = 1, we have BCR(1*2^m-1) = 2*0 = 0 only for m >= 1, because, for m = 0, we have BCR(1*2^0-1) = BCR(0) = 1 <> 2*0 = 0.
%F A345401 a(n) = BCR(2*n) + 1 for n >= 1.
%F A345401 a(n) = 2*A059894(n) + 1 for n >= 1. - _Hugo Pfoertner_, Jun 18 2021
%e A345401 a(0) = 1 because BCR(1*2^m-1) = 2*0 = 0 for m >= 1 (A000225).
%e A345401 a(1) = 3 because BCR(3*2^m-1) = 2*1 = 2 for m >= 0 (A153893).
%e A345401 a(2) = 7 because BCR(7*2^m-1) = 2*2 = 4 for m >= 0 (A086224).
%e A345401 Indeed, a(1) = 3 because 3*2^m-1 = 1011..11_2 (i.e., 10 followed by m 1's), whose bit complement is 0100..00, which reverses to 10_2 = 2 = 2*1.
%e A345401 Also, a(43) = 75 because 75*2^m-1 = 100101011..11_2 (i.e., 1001010 followed by m 1's), whose bit complement is 011010100..00, which reverses to 1010110_2 = 86 = 2*43.
%Y A345401 Cf. A036044 (BCR), A059894.
%Y A345401 When BCR(n) = 0, 2, 4, 6, 8, 10, 12, then corresponding a(n) = h = 1, 3, 7, 5, 15, 11, 13 and numbers h*2^m-1 are respectively in A000225, A153893, A086224, A153894, A196305, A086225, A198274.
%K A345401 nonn,base,easy
%O A345401 0,2
%A A345401 _Bernard Schott_, Jun 18 2021