%I #20 Jul 07 2022 20:29:05

%S 0,1,2,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,19,20,20,22,23,24,25,26,

%T 27,28,29,30,30,32,33,34,35,36,37,38,39,40,40,42,43,44,45,46,47,48,49,

%U 50,50,52,53,54,55,56,57,58,59,60,60,62,63,64,65,66,67,68,69,70,70,72,73,74,75,76,77,78,79,80,80

%N a(n) is the number of digits in the concatenation of a(0) to a(n-1) that are equal to the corresponding digit in the concatenation of all integers >= 0, with a(0) = 0.

%C Let 'G' be the infinite string of numbers formed by the concatenation of all integers >= 0. Under the first '0' write '0', and from then on write the number of digits so far written that equal the corresponding digit in G. The numbers thus written form this sequence.

%C For the first 10 million terms the largest single increment occurs after a(166308) = 52903, and all 5 digits of 52903 match the digits in G, thus a(166309) = 52908. An increment of 5 also occurs after 190 other terms. The longest run of terms which do not change starts at a(112962) = 23232, and 23232 continues for another 87 terms. Although '2' and '3' appear in G during this run it so happens that they are in the alternating position to the '2' and '3' in 23232. This raises the question of whether the longest possible run of non-changing terms is finite or unbounded as n goes to infinity. a(10000000) = 1792311.

%e a(1) = 1 as a(0) = 0, and 0 equals the corresponding digit 0 in G.

%e a(10) = 10 as a(0) = 0, a(1) = 1, ... a(9) = 9, and the digits in all entries a(0) to a(9) equal the corresponding digit in G.

%e a(11) = 12 as a(10) = 10, and the digits '1' and '0' from '10' equal the corresponding digits '1' and '0' in G, thus a(11) = a(10) + 2 = 12.

%e a(12) = 13 as a(11) = 12, and the digit '1' from '12' equals the corresponding digit '1' in G. However the digit '2' in '12' corresponds to the second digit of '11' in G. Thus a(12) = a(11) + 1 = 13.

%Y Cf. A007908, A007376, A117804.

%K nonn,base

%O 0,3

%A _Scott R. Shannon_, Jan 11 2020