A367307 Reverse bits in blocks in binary expansion of n where blocks are separated by every second 1 bit starting from the most significant 1 bit as the first separator.

0, 1, 2, 3, 4, 6, 5, 7, 8, 12, 10, 13, 9, 11, 14, 15, 16, 24, 20, 25, 18, 21, 26, 27, 17, 19, 22, 23, 28, 30, 29, 31, 32, 48, 40, 49, 36, 41, 50, 51, 34, 37, 42, 43, 52, 54, 53, 55, 33, 35, 38, 39, 44, 46, 45, 47, 56, 60, 58, 61, 57, 59, 62, 63, 64, 96, 80, 97

Offset: 0

Author

Mikhail Kurkov, Nov 13 2023

Keywords

• nonn
• base
• easy

Comments

Self-inverse permutation of nonnegative integers.

Examples

			For n = 100930, reversals are each (...) block
  n    = 100930 = binary 1(1000)1(0100)1(000010)
  a(n) = 72016  = binary 1(0001)1(0010)1(010000)
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192
• Index entries for sequences that are permutations of the nonnegative integers

Crossrefs

Cf. A053644, A053645, A059893, A333776, A333777.

Programs

PARI

a(n) = my(v1); v1 = binary(n); my(A = 1); while(A <= #v1, my(B = A, C = 0, D); A++; while(A <= #v1, C += v1[A]; if(v1[A] && (C == 1), D = A); if(C < 2, A++, break)); if(C > 0, v1[D] = 0; v1[A + B - D] = 1)); fromdigits(v1, 2)

Formula

a(n) = b(c(f(n)) + g(n)) = b(h(c(n))) for n > 0 with a(0) = 0 where

b(n) = b(f(h(n))) + g(n) for n > 0 with b(0) = 0,

c(n) = h(c(f(n)) + g(n)) for n > 0 with c(0) = 0,

f(n) = A053645(n), g(n) = A053644(n) and where h(n) = A059893(n).

Note that c (A333776) is the inverse permutation to b (A333777).

a(n) < 2^k iff n < 2^k for k >= 0.