A339723 - OEIS

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.

%I #21 Dec 18 2020 04:16:48

%S 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,

%T 21,31,24,25,28,32,33,34,35,36,54,45,59,37,55,38,53,47,39,40,41,61,42,

%U 63,43,62,44,58,46,48,49,50,51,52,56,57,60,64,65,66,67,68,69

%N 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.

%H Finnegan R. Manthe, <a href="https://onlinegdb.com/Bk3erYSnw">Generation Code</a>

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>

%e 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.

%e 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.

%Y Cf. A175930 (concatenated run lengths).

%K nonn,base

%O 1,2

%A _Finnegan R. Manthe_, Dec 14 2020