%I #10 Jun 10 2023 13:47:28

%S 1,1,1,1,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,2,5,2,2,2,2,2,2,1,1,

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

%N Number of contiguous digits i in the counting numbers, for i=2.

%C Write the natural numbers as an infinite sequence of digits, starting at the left; and count the digits of each group of "2"'s you encounter (the smallest "groups" have one "2").

%H Harvey P. Dale, <a href="/A103756/b103756.txt">Table of n, a(n) for n = 1..1000</a>

%e a(1)=1: the first "2" of the counting numbers, isolated.

%e a(2)=1: the second "2", second digit of number 12, isolated.

%e a(3)=1 the first digit of number 20, isolated.

%e a(4)=1: the first digit of number 21, isolated.

%e a(5)=3: two "2"'s of 22 followed by one "2" of 23 = three "2"'s, the next group of "2"'s.

%t Length/@Select[Split[Flatten[IntegerDigits/@Range[300]]],Union[#]=={2}&] (* _Harvey P. Dale_, Jun 10 2023 *)

%K nonn,base

%O 1,5

%A _Alexandre Wajnberg_, Mar 28 2005