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).

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, 23, 31, 19, 27, 17, 32, 48, 40, 56, 36, 52, 44, 60, 42, 58, 38, 54, 46, 62, 34, 50, 49, 41, 57, 37, 53, 45, 61, 43, 59, 39, 55, 47, 63, 35, 51, 33, 64, 96, 80, 112

Offset

1,2

Comments

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).

Links

Table of n, a(n) for n=1..67.

Peter J. Taylor, Permutation of the natural numbers from operation related to binary expansion of n, answer to question on MathOverflow (2023).

Index entries for sequences that are permutations of the natural numbers

Index entries for sequences related to binary expansion of n

Prog

(PARI) b(n)=my(L=logint(n, 2), A=n); for(i=0, L, A+=2^i*bittest(A, L-i)); A;
a(n)=(b(2*n) - 1)/2 - n

Crossrefs

Cf. A000523, A004754, A059893.

Sequence in context: A073208 A182113 A361647 * A026236 A238862 A048201

Adjacent sequences: A354898 A354899 A354900 * A354902 A354903 A354904

Keyword

nonn

Author

Mikhail Kurkov, Jun 11 2022

Status

approved