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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
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
0,2
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.
FORMULA
a(n) = BCR(2*n) + 1 for n >= 1.
a(n) = 2*A059894(n) + 1 for n >= 1. - Hugo Pfoertner, Jun 18 2021
EXAMPLE
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.
Sequence in context: A184162 A369277 A324821 * A059912 A354008 A115765
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Jun 18 2021
STATUS
approved