A345401 a(n) is the unique odd number h such that BCR(h*2^m-1) = 2n (except for BCR(0) = 1) where BCR is bit complement and reverse per A036044.

1, 3, 7, 5, 15, 11, 13, 9, 31, 23, 27, 19, 29, 21, 25, 17, 63, 47, 55, 39, 59, 43, 51, 35, 61, 45, 53, 37, 57, 41, 49, 33, 127, 95, 111, 79, 119, 87, 103, 71, 123, 91, 107, 75, 115, 83, 99, 67, 125, 93, 109, 77, 117, 85, 101, 69, 121, 89, 105, 73, 113, 81, 97, 65, 255, 191

Offset: 0

Author

Bernard Schott, Jun 18 2021

Keywords

• nonn
• base
• easy

Comments

This sequence is a permutation of the odd numbers.

We have BCR(a(n)*2^m-1) = 2n when n = 0 for m >= 1, and BCR(a(n)*2^m-1) = 2n when n >= 1 for m >= 0.

Why this exception when n = 0? As a(0) = 1, we have BCR(1*2^m-1) = 2*0 = 0 only for m >= 1, because, for m = 0, we have BCR(1*2^0-1) = BCR(0) = 1 <> 2*0 = 0.

Examples

			a(0) = 1 because BCR(1*2^m-1) = 2*0 = 0 for m >= 1 (A000225).
a(1) = 3 because BCR(3*2^m-1) = 2*1 = 2 for m >= 0 (A153893).
a(2) = 7 because BCR(7*2^m-1) = 2*2 = 4 for m >= 0 (A086224).
Indeed, a(1) = 3 because 3*2^m-1 = 1011..11_2 (i.e., 10 followed by m 1's), whose bit complement is 0100..00, which reverses to 10_2 = 2 = 2*1.
Also, a(43) = 75 because 75*2^m-1 = 100101011..11_2 (i.e., 1001010 followed by m 1's), whose bit complement is 011010100..00, which reverses to 1010110_2 = 86 = 2*43.
		

Crossrefs

Cf. A036044 (BCR), A059894.

When BCR(n) = 0, 2, 4, 6, 8, 10, 12, then corresponding a(n) = h = 1, 3, 7, 5, 15, 11, 13 and numbers h*2^m-1 are respectively in A000225, A153893, A086224, A153894, A196305, A086225, A198274.

Formula

a(n) = BCR(2*n) + 1 for n >= 1.

a(n) = 2*A059894(n) + 1 for n >= 1. - Hugo Pfoertner, Jun 18 2021