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a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.
5

%I #24 Nov 30 2019 02:05:24

%S 1,0,2,3,11,71,29,9,683,67,7,743,739,1933,23,161,21,37,19,17,119,49,

%T 332534262883,13,39,13739483941387,83,111,79853560395691,5431567,

%U 70610371,69,51,4112497,28384496881337963,353,77,1531,42787,63,27,41,709,33,81,487,139697

%N a(1) = 1, a(2) = 0; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.

%H Chai Wah Wu, <a href="/A241811/b241811.txt">Table of n, a(n) for n = 1..68</a>

%e a(1)=1 and a(2)=0. a(1) U a(2) = 10 and its divisors are 1, 2, 5, 10. Therefore 2 is the least number not yet present in the sequence which divides 10. Again, a(1) U a(2) U a(3) = 102 and its divisors are 1, 2, 3, 6, 17, 34, 51, 102. Therefore a(4)=3, etc.

%p with(numtheory):

%p T:=proc(t) local x, y; x:=t; y:=0; while x>0 do x:=trunc(x/10); y:=y+1; od; end:

%p P:=proc(q) local a,b,c,k,n; b:=10; print(1); print(0); c:=[0,1];

%p for n from 1 to q do a:=sort([op(divisors(b))]); for k from 2 to nops(a) do

%p if not member(a[k],c) then c:=[op(c),a[k]]; b:=a[k]+b*10^T(a[k]); print(a[k]); break;

%p fi; od; od; end: P(30);

%Y Cf. A096097, A096098, A240588.

%K nonn,base

%O 1,3

%A _Paolo P. Lava_, Apr 29 2014

%E a(23)-a(28) from _Zak Seidov_, May 08 2014

%E a(29)-a(47) from _Giovanni Resta_, Aug 15 2019