%I #55 Sep 23 2016 10:42:00

%S 0,0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,0,10,1,2,1,3,1,4,1,5,1,6,1,7,1,

%T 8,1,9,1,10,2,2,3,2,4,2,5,2,6,2,7,2,8,2,9,2,10,3,3,4,3,5,3,6,3,7,3,8,

%U 3,9,3,10,4,4,5,4,6,4,7,4,8,4,9,4,10,5,5,6,5,7,5,8,5,9,5,10,6,6,7,6,8,6,9,6,10,7,7,8,7,9,7,10

%N a(n) is the number of times that a(n-1) appears in the concatenation of all numbers from a(0) to a(n-2), with a(0) = 0.

%C 138, 185 and 199 are three smallest numbers that do not appear among the first 5000 terms of the sequence. They first appear at n = 8776, 5117 and 10580 respectively.

%C 1187 and 1190 are two smallest numbers that do not appear among the first 100000 terms.

%C Question: will every natural number eventually appear in the sequence?

%C The sequence can be started with any number a(0). The terms will be different, but for larger n behavior will be similar for all a(0).

%H Yuriy Sibirmovsky, <a href="/A276457/b276457.txt">Table of n, a(n) for n = 0..100000</a>

%H Yuriy Sibirmovsky, <a href="/A276457/a276457.png">Plot for n=0..4999</a>

%H Yuriy Sibirmovsky, <a href="/A276457/a276457_1.png">Plot for n=0..59999</a>

%F a(n) = A142150(n) = A171181(n), if 0<=n<=20.

%F a(n) = A248034(n-19), if 21<=n<=120. - _Omar E. Pol_, Sep 03 2016

%e From a(0) to a(0), a(1) appears once, thus a(2) = 1.

%e From a(0) to a(1), a(2) appears 0 times, thus a(3) = 0.

%e ...

%e From a(0) to a(19), a(20) = 10 appears once, in the form of '1,0'. Thus a(21) = 1.

%t Nm=100;

%t A=Table[0,{j,1,Nm}];

%t A[[3]]=1;

%t Do[B=Table[IntegerDigits[A[[l]]],{l,1,j-1}];

%t A[[j+1]]=SequenceCount[Flatten[B],IntegerDigits[A[[j]]]],{j,3,Nm-1}];

%t A

%Y Cf. A142150, A171181, A248034.

%Y Similar in spirit to van Eck's A181391.

%K nonn,base,look,hear

%O 0,5

%A _Yuriy Sibirmovsky_, Sep 03 2016