A345352 The binary expansion of a(n) is obtained by left-padding the binary expansion of n with 0's so that it has 2^k digits for some k > 0 as small as possible and then reversing the first half and the second half of this binary expansion.

0, 1, 2, 3, 8, 10, 9, 11, 4, 6, 5, 7, 12, 14, 13, 15, 128, 136, 132, 140, 130, 138, 134, 142, 129, 137, 133, 141, 131, 139, 135, 143, 64, 72, 68, 76, 66, 74, 70, 78, 65, 73, 69, 77, 67, 75, 71, 79, 192, 200, 196, 204, 194, 202, 198, 206, 193, 201, 197, 205

Offset

0,3

Comments

This sequence is a self-inverse permutation of the nonnegative integers with infinitely many fixed points (A345362).

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Index entries for sequences related to binary expansion of n

Index entries for sequences that are permutations of the natural numbers

Formula

A000120(a(n)) = A000120(n).

a(A001196(n)) = A001196(a(n)).

a(n) < 2^2^k for any n < 2^2^k.

a(2^k) = 2^A054429(k) for any k > 0.

Example

For n = 43:
- the binary expansion of n is "101011",
- it has 6 binary digits, so we pad it with 2 leading 0's,
- the first half is "0010" and its reversal is "0100",
- the second half is "1011" and its reversal is "1101",
- so the binary expansion of a(43) is "01001101",
- and a(43) = 77.

Prog

(PARI) a(n) = { my (b=binary(n), x); for (k=1, oo, x=2^k-#b; if (x>=0, b=concat(vector(x), b); return (fromdigits(concat(Vecrev(b[1..#b/2]), Vecrev(b[#b/2+1..#b])), 2)))) }
(Python)
def a(n):
    b = bin(n)[2:]
    bb = bin(len(b))[2:]
    if bb != '1' + '0'*(len(bb)-1): b = '0'*(2**len(bb) - len(b)) + b
    return int(b[:len(b)//2][::-1] + b[len(b)//2:][::-1], 2)
print([a(n) for n in range(60)]) # Michael S. Branicky, Jun 15 2021

Crossrefs

Cf. A000120, A001196, A054429, A345362 (fixed points).

Sequence in context: A019307 A342849 A247580 * A028742 A028771 A023709

Adjacent sequences: A345349 A345350 A345351 * A345353 A345354 A345355

Keyword

nonn,base

Author

Rémy Sigrist, Jun 15 2021

Status

approved