login
Consider the concatenation of a(n) and a(n+1) as the integer A; consider the concatenation of the last digit of a(n) and the first digit of a(n+1) as the integer B. Now A is divisible by B.
1

%I #9 Mar 23 2018 01:57:19

%S 1,2,3,4,5,6,7,8,9,10,11,28,71,17,111,24,19,110,12,18,48,72,50,13,68,

%T 73,16,38,151,14,88,74,39,77,106,95,76,26,91,20,15,37,92,40,22,31,33,

%U 82,34,78,165,36,112,84,64,80,44,55,131,30,60,66,93,86,94,136,62,118,179,216,123,42,32,109,204

%N Consider the concatenation of a(n) and a(n+1) as the integer A; consider the concatenation of the last digit of a(n) and the first digit of a(n+1) as the integer B. Now A is divisible by B.

%C The sequence starts with a(1) = 1 and is always extended with the smallest integer not yet present in the sequence that does not lead to a contradiction.

%C Leading zeros must be erased from the divisor: the pair [10,11] is thus accepted because 1011 is divisible by 1.

%C Lars Blomberg (in a private mail):

%C ...The integer 21 is never used because:

%C ...Let the previous number be x*10 + y, the new number is 21.

%C ...Then x*1000 + y*100 + 21 (which is odd) must be divisible by y*10+2

%C ...which is even, not possible.

%C ...The same goes for 23, 25, 27, 29 and 41, 43, 45, 47, 49 and many more.

%H Lars Blomberg, <a href="/A300819/b300819.txt">Table of n, a(n) for n = 1..10000</a>

%e The concatenation [12] is divisible by 12, of course. A similar situation applies for the pairs [23], [34], [45], etc. When it comes to [910] we see that 910 is divisible by 91 (concatenation of 9 and 1, which are the last and first digits of the pair [9,10]). As we consider that [1011] is divisible by 1 (see the comment section), we accept that the integer 11 comes immediately after 10. But 12 cannot follow now, as [1112] is not divisible by 11. We have then to wait until 28 shows up as [1128] is divisible by 12 (quotient = 94). Etc.

%K nonn,base

%O 1,2

%A _Eric Angelini_ and _Lars Blomberg_, Mar 13 2018