A269167 -id:A269167 - cp's OEIS frontend

A269168 Rule 30 binary tree permutation: a(1) = 1, a(2n) = A269160(a(n)), a(2n+1) = A269164(1+a(n)).

1, 7, 2, 25, 9, 14, 3, 111, 33, 63, 11, 50, 18, 13, 4, 401, 143, 231, 41, 193, 79, 53, 15, 222, 66, 126, 22, 51, 17, 28, 5, 1783, 529, 945, 175, 825, 295, 223, 55, 839, 257, 497, 95, 203, 69, 49, 19, 802, 286, 462, 82, 386, 158, 106, 30, 221, 67, 119, 21, 100, 36, 27, 6, 6409, 2295, 3703, 657, 3159, 1201, 849, 233

Offset: 1

Antti Karttunen, Feb 21 2016

This sequence can be represented as a binary tree. Each left hand child is produced as A269160(n), and each right hand child as A269164(1+n), when the parent node contains n:
                                     |
                  ...................1...................
                 7                                       2
      25......../ \........9                  14......../ \........3
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  111      33         63       11         50       18         13       4
401 143  231 41    193  79   53  15    222  66  126  22     51  17   28 5
etc.
Each maximal leftward branch (e.g. 1, 7, 25, ... (= A110240) or 9, 63, 193, ... or 2, 14, 50, ...) gives a trajectory of Rule 30 cellular automaton starting from a particular "seed configuration" which are given in A269164.

Antti Karttunen, Table of n, a(n) for n = 1..1023
Antti Karttunen, Entanglement Permutations, 2016-2017
Index entries for sequences that are permutations of the natural numbers

Inverse: A269167.
Cf. A269160, A269164.
Cf. A110240 (the left edge).

nmax = (* sequence length *) 100; terms (* from A269164 *) = 2000; Clear[a, f]; A269160[n_] := BitXor[n, BitOr[2 n, 4 n]]; f[max_] := f[max] = (s = Sort[Table[A269160[n], {n, 0, max}]]; Complement[Range[Last[s]], s][[1 ;; terms]]); f[terms]; f[max = 2 terms]; While[f[max] != f[max/2], max = 2 max]; A269164[n_Integer] := If[n > Length[f[max]], 0, f[max][[n]]]; a[1] = 1; a[n_] := a[n] = If[EvenQ[n], A269160[a[n/2]], A269164[1 + a[(n - 1)/2]]]; A269168 = Table[a[n], {n, 1, nmax}] (* Jean-François Alcover, Feb 23 2016 *)

a(1) = 1, after which, a(2n) = A269160(a(n)), a(2n+1) = A269164(1+a(n)).

A269169 The least monotonic left inverse for A269164.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 41, 41, 41, 41, 41, 41, 42, 42, 43, 44, 45, 46, 47, 48, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66

Offset: 1

Antti Karttunen, Feb 21 2016

nonn

a(n) = number of terms of A269164 <= n.

Antti Karttunen, Table of n, a(n) for n = 1..8192

Cf. A269164, A269167.

terms = 100; Clear[f]; f[max_] := f[max] = (s = Sort[Table[BitXor[n, BitOr[ 2 n, 4 n]], {n, 0, max}]]; Complement[Range[Last[s]], s][[1 ;; terms]]); f[terms]; f[max = 2 terms]; While[Print[max]; f[max] != f[max/2], max = 2 max]; A269164 = f[max]; a[n_] := Count[A269164, k_ /; k <= n]; Table[ a[n], {n, 1, Length[A269164]}] (* Jean-François Alcover, Feb 23 2016 *)

Other identities. For all n >= 1:

a(A269164(n)) = n.

cp's OEIS Frontend

A269168 Rule 30 binary tree permutation: a(1) = 1, a(2n) = A269160(a(n)), a(2n+1) = A269164(1+a(n)).

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