A346306 Position in A076478 of the binary complement of the n-th word in A076478.
2, 1, 6, 5, 4, 3, 14, 13, 12, 11, 10, 9, 8, 7, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 126, 125, 124, 123
Offset: 1
Examples
The first fourteen words w(n) are 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, so that a(3) = 6. From _Paolo Xausa_, Mar 09 2023: (Start) Written as an irregular triangle, where row r >= 1 has length 2^r and row sum is A103897(r), the sequence begins: 2, 1; 6, 5, 4, 3; 14, 13, 12, 11, 10, 9, 8, 7; 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15; ... (End)
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..16382 (for all words with length <= 13)
- Index entries for sequences that are permutations of the natural numbers
Programs
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Mathematica
(See A007931.) A346306[rowmax_]:=Table[Range[2^(r+1)-2,2^r-1,-1],{r,rowmax}]; A346306[6] (* Paolo Xausa, Mar 09 2023 *)
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Python
from itertools import product def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z}) def wgen(maxdigits): for digits in range(1, maxdigits+1): for b in product("01", repeat=digits): yield "".join(b) def auptod(maxdigits): w = [None] + [wn for wn in wgen(maxdigits)] return [w.index(comp(w[n])) for n in range(1, 2**(maxdigits+1) - 1)] print(auptod(6)) # Michael S. Branicky, Sep 03 2021
Formula
a(n) = 3*(2^d - 1) - n, where 2^d - 1 <= n <= 2^(d+1) - 2. - Michael S. Branicky, Sep 03 2021
Comments