A322010 Inverse permutation to A322000.

0, 1, 2, 4, 6, 10, 14, 20, 26, 36, 3, 5, 7, 11, 15, 21, 27, 37, 46, 59, 8, 12, 16, 22, 28, 38, 47, 60, 72, 90, 17, 23, 29, 39, 48, 61, 73, 91, 108, 130, 30, 40, 49, 62, 74, 92, 109, 131, 152, 182, 50, 63, 75, 93, 110, 132, 153, 183, 212, 248, 76, 94, 111, 133, 154, 184, 213, 249

Offset: 0

Author

M. F. Hasler, Feb 19 2019

Keywords

• nonn

Comments

a(n) is the position of n in the list A322000 of "decibinary numbers", i.e., integers sorted according to their decibinary value A028897(n) = Sum d[i]*2^i, where d[i] are the decimal digits of n.

For 0 <= m <= 9, we have a(n) = A322003(n) = A000123(n-1), because 1..9 are the first few terms of A322000 where the decibinary value increases.

We see that a(10..19) = a(2..9)+1 concatenated with (46, 49). Then, a(20..29) = a(12..19)+1 concatenated with (72, 90). Then, a(30..39) = a(22..29)+1 concatenated with (108, 130), and so on. This yields an alternate way to compute the sequence.

Crossrefs

Cf. A322000, A028897, A322003, A072170(.,10), A000123.

Programs

PARI

vec_A322010=vecsort(A,,1)[1..vecmin(setminus([1..#A],Set(A)))-1] \\ Assumes the vector A = A322000(1..N) has been computed for some N. Exclude initial 0's to have correct (1-based) indices of the vectors.