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Inverse permutation to A322000.
0

%I #10 Mar 22 2022 10:01:46

%S 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,

%T 47,60,72,90,17,23,29,39,48,61,73,91,108,130,30,40,49,62,74,92,109,

%U 131,152,182,50,63,75,93,110,132,153,183,212,248,76,94,111,133,154,184,213,249

%N Inverse permutation to A322000.

%C 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.

%C 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.

%C 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.

%o (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.

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

%K nonn

%O 0,3

%A _M. F. Hasler_, Feb 19 2019