login
A209862
Permutation of nonnegative integers which maps A209642 into ascending order (A209641).
22
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 15, 16, 17, 18, 20, 24, 19, 21, 25, 22, 26, 28, 23, 27, 29, 30, 31, 32, 33, 34, 36, 40, 48, 35, 37, 41, 49, 38, 42, 50, 44, 52, 56, 39, 43, 51, 45, 53, 57, 46, 54, 58, 60, 47, 55, 59, 61, 62, 63, 64, 65, 66, 68, 72, 80, 96, 67, 69, 73, 81, 97, 70, 74, 82, 98, 76, 84, 100, 88, 104, 112, 71, 75, 83
OFFSET
0,3
COMMENTS
Conjecture: For all n, a(A054429(n)) = A054429(a(n)), i.e. A054429 acts as a homomorphism (automorphism) of the cyclic group generated by this permutation. This implies also a weaker conjecture given in A209860.
From Gus Wiseman, Aug 24 2021: (Start)
As a triangle with row lengths 2^n, T(n,k) for n > 0 appears (verified up to n = 2^15) to be the unique nonnegative integer whose binary indices are the k-th subset of {1..n} containing n. Here, a binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion, and sets are sorted first by length, then lexicographically. For example, the triangle begins:
1
2 3
4 5 6 7
8 9 10 12 11 13 14 15
16 17 18 20 24 19 21 25 22 26 28 23 27 29 30 31
Mathematica: Table[Total[2^(Append[#,n]-1)]&/@Subsets[Range[n-1]],{n,5}]
Row lengths are A000079 (shifted right). Also Column k = 1.
Row sums are A010036.
Using reverse-lexicographic order gives A059893.
Using lexicographic order gives A059894.
Taking binary indices to prime indices gives A339195 (or A019565).
The ordering of sets is A344084.
A version using Heinz numbers is A344085.
(End)
FORMULA
EXAMPLE
From Gus Wiseman, Aug 24 2021: (Start)
The terms, their binary expansions, and their binary indices begin:
0: ~ {}
1: 1 ~ {1}
2: 10 ~ {2}
3: 11 ~ {1,2}
4: 100 ~ {3}
5: 101 ~ {1,3}
6: 110 ~ {2,3}
7: 111 ~ {1,2,3}
8: 1000 ~ {4}
9: 1001 ~ {1,4}
10: 1010 ~ {2,4}
12: 1100 ~ {3,4}
11: 1011 ~ {1,2,4}
13: 1101 ~ {1,3,4}
14: 1110 ~ {2,3,4}
15: 1111 ~ {1,2,3,4}
(End)
KEYWORD
nonn
AUTHOR
Antti Karttunen, Mar 24 2012
STATUS
approved