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A322010
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Inverse permutation to A322000.
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0
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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
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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PROG
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(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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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