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A354901 a(n) = (b(2n) - 1)/2 - n for n > 0. To get b(n) start with A = n and then for i = 0..f(n) apply A := A + 2^i*T(A, f(n) - i) where T(n, k) = floor(n/2^k) mod 2 and f(n) = A000523(n).

%I A354901 #28 Nov 06 2024 04:55:19
%S A354901 1,2,3,4,6,7,5,8,12,10,14,13,11,15,9,16,24,20,28,22,30,18,26,25,21,29,
%T A354901 23,31,19,27,17,32,48,40,56,36,52,44,60,42,58,38,54,46,62,34,50,49,41,
%U A354901 57,37,53,45,61,43,59,39,55,47,63,35,51,33,64,96,80,112
%N A354901 a(n) = (b(2n) - 1)/2 - n for n > 0. To get b(n) start with A = n and then for i = 0..f(n) apply A := A + 2^i*T(A, f(n) - i) where T(n, k) = floor(n/2^k) mod 2 and f(n) = A000523(n).
%C A354901 Subsequences from a(2^m) to a(2^(m+1) - 1) for m >= 0 contain all numbers k such that 2^m <= k < 2^(m+1). This fact was proved (see _Peter J. Taylor_ link).
%H A354901 Peter J. Taylor, <a href="https://mathoverflow.net/a/455313/231922">Permutation of the natural numbers from operation related to binary expansion of n</a>, answer to question on MathOverflow (2023).
%H A354901 <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%H A354901 <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%o A354901 (PARI) b(n)=my(L=logint(n,2),A=n); for(i=0,L, A+=2^i*bittest(A,L-i)); A;
%o A354901 a(n)=(b(2*n) - 1)/2 - n
%Y A354901 Cf. A000523, A004754, A059893.
%K A354901 nonn
%O A354901 1,2
%A A354901 _Mikhail Kurkov_, Jun 11 2022