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

Author

Rémy Sigrist, Jun 15 2021

Keywords

• nonn
• base

Comments

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

Examples

			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.
		

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

Crossrefs

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

Programs

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

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.