|
|
A339723
|
|
Start with a(1)=1. Thereafter, to get a(n), write a(n-1) in binary, write down all its run lengths, and concatenate them in binary, getting k. Then a(n) = k unless k is already in the sequence, in which case a(n) = smallest missing number.
|
|
0
|
|
|
1, 2, 3, 4, 6, 5, 7, 8, 9, 13, 11, 14, 10, 15, 12, 16, 17, 18, 27, 22, 29, 19, 26, 23, 20, 30, 21, 31, 24, 25, 28, 32, 33, 34, 35, 36, 54, 45, 59, 37, 55, 38, 53, 47, 39, 40, 41, 61, 42, 63, 43, 62, 44, 58, 46, 48, 49, 50, 51, 52, 56, 57, 60, 64, 65, 66, 67, 68, 69
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
LINKS
|
|
|
EXAMPLE
|
for n=3, a(2) = 2 in binary is 10 with run length 1,1, concatenated is 11 which in base 10 is k=3. So a(3)=3.
For n=7 doing this gives 7 which is already in the sequence (at n=5) so we put 6 as it is the smallest number not already in the sequence.
|
|
CROSSREFS
|
Cf. A175930 (concatenated run lengths).
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|