|
|
A119435
|
|
a(n) = (binary reversal of n)-th integer among those positive integers not occurring earlier in the sequence.
|
|
9
|
|
|
1, 2, 5, 3, 9, 7, 13, 4, 17, 12, 23, 10, 22, 18, 29, 6, 33, 24, 43, 16, 40, 31, 51, 14, 41, 30, 53, 25, 49, 38, 61, 8, 65, 45, 83, 32, 76, 58, 95, 21, 74, 55, 94, 42, 87, 68, 107, 19, 78, 56, 100, 39, 91, 70, 113, 34, 89, 66, 112, 52, 104, 81, 125, 11, 129, 86, 163, 60, 148
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
This sequence is a permutation of the positive integers.
[Proof from N. J. A. Sloane, Apr 20 2022: a(n) always exists, so the sequence is infinite. Every time n is a power of 2, n-reversed is 1, and a(n) is the smallest missing number. Since there are infinitely many powers of 2, every number will eventually appear.]
|
|
LINKS
|
|
|
EXAMPLE
|
12 in binary is 1100; so its binary reversal is 0011, which is 3 in decimal. Those positive integers not among the first 11 terms of the sequence are 6,8,10,11,14,..., and the third of these is 10, so a(12) = 10.
|
|
MATHEMATICA
|
Block[{a = {1}, nn = 69}, Do[AppendTo[a, #] &@ Complement[Range[i + 2 nn], #][[FromDigits[#, 2] &@ Reverse@ IntegerDigits[i, 2]]] &@ a, {i, 2, nn}]; a] (* Michael De Vlieger, Sep 03 2017 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|