A367306 - OEIS

A367306

Move bits in blocks in binary expansion of n where blocks are defined in Comments.

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 9, 11, 12, 13, 14, 15, 16, 20, 17, 21, 18, 19, 22, 23, 24, 26, 25, 27, 28, 29, 30, 31, 32, 40, 33, 41, 34, 35, 42, 43, 36, 37, 38, 39, 44, 45, 46, 47, 48, 52, 49, 53, 50, 51, 54, 55, 56, 58, 57, 59, 60, 61, 62, 63, 64, 80, 65, 81

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OFFSET

0,3

COMMENTS

Define the "blocks" of the binary expansion of n as runs of 0's followed by a 1 bit and additionally followed with a 0 bit if it lies immediately after the 1 bit. For example, the blocks in 45986 (shown in parentheses) are 1(01)1(001)11(010)(0010). To obtain binary expansion of a(n) from binary expansion of n, move the rightmost bit in each block to the position after leftmost bit in the same block and leave all other 1 bits unchanged (see the Example).

LINKS

Table of n, a(n) for n=0..67.

Index entries for sequences that are permutations of the nonnegative integers

FORMULA

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

Conjecture: a(n) = A200714(p(n) + 1) where p(n) is an inverse permutation to A358654.

EXAMPLE

For n = 45986, moves are in each (...) block

n = 45986 = binary 1(01)1(001)11(010)(0010)

a(n) = 46481 = binary 1(01)1(010)11(001)(0001)

We have moves in last 3 blocks, while in the first block the rightmost 1 bit is already lies on the position after leftmost bit in block, so we have no change in this case.

PROG

(PARI) a(n) = my(v1); v1 = binary(n); my(A = 1); while(A <= #v1, while(A <= #v1 && v1[A], A++); my(B = A); while((A + 1) <= #v1 && !v1[A + 1], A++); if(A < #v1, if((A + 2) <= #v1 && !v1[A + 2], B = A; A++); v1[A + 1] = !v1[A + 1]; v1[B + 1] = !v1[B + 1]); A += 2); fromdigits(v1, 2)

CROSSREFS

Cf. A200714, A358654.

Sequence in context: A353907 A363271 A031044 * A194978 A195096 A194977

Adjacent sequences: A367303 A367304 A367305 * A367307 A367308 A367309

KEYWORD

nonn,base

AUTHOR

Mikhail Kurkov, Nov 13 2023

STATUS

approved