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Consider the runs of 1's in the binary representation of n, each of these runs being on the edge of the binary representation n and/or being bounded by 0's. a(n) = the length of the shortest such run of 1's in binary n.
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%I #13 Oct 27 2017 20:59:56

%S 1,1,2,1,1,2,3,1,1,1,1,2,1,3,4,1,1,1,1,1,1,1,1,2,1,1,2,3,1,4,5,1,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,2,1,1,2,2,3,1,1,2,4,1,5,6,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,2,1,1,2,2,1,1

%N Consider the runs of 1's in the binary representation of n, each of these runs being on the edge of the binary representation n and/or being bounded by 0's. a(n) = the length of the shortest such run of 1's in binary n.

%H Michael De Vlieger, <a href="/A144790/b144790.txt">Table of n, a(n) for n = 1..16384</a>

%e 19 in binary is 10011. The runs of 1's are as follows: (1)00(11). The shortest of these runs contains exactly one 1. So a(19) = 1.

%t Array[Min@ Map[Length, Select[Split@ IntegerDigits[#, 2], First@ # == 1 &]] &, 105] (* _Michael De Vlieger_, Oct 26 2017 *)

%Y Cf. A038374, A144789.

%K base,nonn

%O 1,3

%A _Leroy Quet_, Sep 21 2008

%E Extended by _Ray Chandler_, Nov 04 2008