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A298847
Lexicographically earliest sequence of distinct positive terms such that, for any n > 0, the number of ones in the binary expansion of n equals one plus the number of zeros in the binary expansion of a(n).
5
1, 3, 2, 7, 5, 6, 4, 15, 11, 13, 9, 14, 10, 12, 8, 31, 23, 27, 19, 29, 21, 22, 17, 30, 25, 26, 18, 28, 20, 24, 16, 63, 47, 55, 39, 59, 43, 45, 35, 61, 46, 51, 37, 53, 38, 41, 33, 62, 54, 57, 42, 58, 44, 49, 34, 60, 50, 52, 36, 56, 40, 48, 32, 127, 95, 111, 79
OFFSET
1,2
COMMENTS
In other words, for any n > 0, A000120(n) = 1 + A023416(a(n)).
This sequence is a self-inverse permutation of the natural numbers, with fixed points A031448.
We can build an analog of this sequence for any base b > 1:
- let s_b be the sum of digits in base b,
- in particular s_2 = A000120 and s_10 = A007953,
- let l_b be the number of digits in base b,
- in particular l_2 = A070939 and l_10 = A055642,
- let f_b be the lexicographically earliest sequence of distinct positive terms such that, for any n > 0, s_b(n) = 1 + (b-1) * l_b(a(n)) - s_b(a(n)),
- in particular, f_2 = a (this sequence),
- f_b is a self-inverse permutation of the natural numbers,
- l_b(n) = l_b(f_b(n)) for any n > 0,
- f_b(b^k) = b^(k+1) - 1 for any k >= 0,
- see also scatterplots of f_3 and f_10 in Links section.
FORMULA
A070939(n) = A070939(a(n)) for any n > 0.
a(2^k) = 2^(k+1) - 1 for any k >= 0.
A000120(n) + A000120(a(n)) = 1 + A070939(n) for any n > 0.
EXAMPLE
The first terms, alongside the binary representations of n and of a(n), are:
n a(n) bin(n) bin(a(n))
-- ---- ------ ---------
1 1 1 1
2 3 10 11
3 2 11 10
4 7 100 111
5 5 101 101
6 6 110 110
7 4 111 100
8 15 1000 1111
9 11 1001 1011
10 13 1010 1101
11 9 1011 1001
12 14 1100 1110
13 10 1101 1010
14 12 1110 1100
15 8 1111 1000
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Jan 27 2018
STATUS
approved