A381839 In the binary expansion of n (without leading zeros): complement the bits strictly between the leftmost and the rightmost 0's, if any.

0, 1, 2, 3, 4, 5, 6, 7, 10, 9, 8, 11, 12, 13, 14, 15, 22, 21, 20, 19, 18, 17, 16, 23, 26, 25, 24, 27, 28, 29, 30, 31, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 47, 54, 53, 52, 51, 50, 49, 48, 55, 58, 57, 56, 59, 60, 61, 62, 63, 94, 93, 92, 91

Offset

0,3

Comments

This sequence is a self-inverse permutation of the nonnegative integers.

This sequence has similarities with A122155 (where we complement bits between the leftmost and the rightmost 1's).

This sequence has infinitely many fixed points: A000225, A030130, and the positive numbers whose binary expansion have exactly two 0's that are also adjacent to each other.

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8191

Index entries for sequences related to binary expansion of n

Index entries for sequences that are permutations of the natural numbers

Formula

a(2*n + 1) = 2*a(n) + 1.

Example

The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     2      10         10
   3     3      11         11
   4     4     100        100
   5     5     101        101
   6     6     110        110
   7     7     111        111
   8    10    1000       1010
   9     9    1001       1001
  10     8    1010       1000
  11    11    1011       1011
  12    12    1100       1100
  13    13    1101       1101
  14    14    1110       1110
  15    15    1111       1111
  16    22   10000      10110

Prog

(PARI) a(n) = { my (b = binary(n)); for (i = 1, #b, if (b[i]==0, forstep (j = #b, 1, -1, if (b[j]==0, for (k = i+1, j-1, b[k] = 1-b[k]; ); return (fromdigits(b, 2)); ); ); ); ); return (n); }
(Python)
def a(n):
    b = bin(n)[2:]
    zl, zr = b.find('0'), b.rfind('0')
    return n if abs(zl-zr) < 2 else int(b[:zl+1]+"".join('0' if bi == '1' else '1' for bi in b[zl+1:zr])+b[zr:], 2)
print([a(n) for n in range(70)]) # Michael S. Branicky, Mar 09 2025
(Python)
def A381839(n): return n^((1<<n.bit_length()-t-1)-1)^(((~n&n+1)<<1)-1) if n and (t:=(s:=bin(n)[2:]).find('0'))!=-1 and s.count('0')>1 else n # Chai Wah Wu, Mar 09 2025

Crossrefs

Cf. A000225, A030130, A122155, A381852.

Sequence in context: A371249 A354369 A199426 * A119257 A376214 A266645

Adjacent sequences: A381836 A381837 A381838 * A381840 A381841 A381842

Keyword

nonn,base,easy

Author

Rémy Sigrist, Mar 08 2025

Status

approved