A261300 - OEIS

%I #31 Apr 21 2024 22:24:32

%S 0,1,2,11,3,21,12,111,4,31,22,211,13,121,112,1111,5,41,32,311,23,221,

%T 212,2111,14,131,122,1211,113,1121,1112,11111,6,51,42,411,33,321,312,

%U 3111,24,231,222,2211,213,2121,2112,21111,15,141,132,1311,123,1221,1212,12111,114,1131,1122,11211,1113,11121,11112,111111,7,61,52

%N Concatenate successive run lengths of 0's in the binary expansion of n, each increased by 1.

%C Any positive number written in binary has its first bit equal to 1. From there on we count the 0 bits up to the next 1 bit or the end of the number. Each count is increased by one because of the impossibility of representing leading 0's in this database.

%C The sequence is prefixed by a conventional a(0)=0, which represents an empty sum or concatenation.

%C The positive integer n written as a binary digit string of length m uniquely decomposes into substrings of '1' followed by a maximal run of '0's where the lengths of these substrings forms a composition of m. a(n) is the concatenation of the parts of the composition of m when written in decimal. See A066099 for the table of composition parts. - _Michael Somos_, Aug 20 2015

%C Suggested by Armands Strazds's sequence A258055.

%F a(n) = Sum_{k=0..f(n)-1} T(n,k)*10^g(n,k) for n > 0 with a(0)=0 where f(n) = A000120(n), T(n,k) = T(floor(n/2),k - n mod 2) for k > 0 with T(n,0) = A001511(n), and g(n,k) = Sum_{j=0..k-1} (1 + floor(log_10(T(n,j)))). - _Mikhail Kurkov_, Nov 25 2019 [verification needed]

%e n=2 is written "10" in binary, so following the initial digit '1', there is one (= 1) bit zero; this 1 becomes increased to yield a(2) = 2.

%e n=3 is written "11" in binary, so following the initial digit '1', there is no (= 0) bit zero; after the next digit '1', there follow again 0 bits '0'. These two 0 are increased to yield two 1's, whence a(3) = 11.

%o (PARI) A261300(n,s="",c=0)={for(i=2,#n=binary(n),c++;n[i]&&s=concat(s,c+c=0));eval(concat(s,c++))}

%Y Cf. A066099, A258055.

%K nonn,base

%O 0,3

%A _M. F. Hasler_, Aug 16 2015