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Start with a(1) = 1 then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(n-1),a(n)).
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%I #8 Dec 10 2014 03:59:35

%S 1,2,4,5,10,20,8,16,25,50,40,32,64,80,100,125,200,160,128,250,400,320,

%T 256,500,625,1000,800,640,512,1024,1250,2000,1280,1600,2500,3125,5000,

%U 3200,2048,2560,4000,6250,10000,6400,4096,5120,8000,10240,8192,12500,15625

%N Start with a(1) = 1 then a(n) = smallest number, not already in the sequence, such that a(n) divides concat(a(n-1),a(n)).

%C Like A249398, but without the constraint a(n) > a(n-1).

%H Paolo P. Lava, <a href="/A249399/b249399.txt">Table of n, a(n) for n = 1..100</a>

%e a(1) = 1;

%e a(2) = 2 -> 12 /2 = 6;

%e Now we cannot use 3 as the next term because it does not divide 23.

%e a(3) = 4 -> 24 / 4 = 6;

%e a(4) = 5 -> 45 / 5 = 9;

%e Again, 3, 6, 7, 8 and 9 cannot be used as the next term.

%e a(5) = 10 -> 510 / 10 = 51;

%e a(6) = 20 -> 1020 / 20 = 51;

%e a(7) = 8 -> 208 / 8 = 26; etc.

%p with(numtheory); P:=proc(q) local a,b,k,n; print(1); a:=1; b:={1};

%p for k from 1 to q do for n from 1 to q do if nops({n} intersect b)<1

%p then if type((a*10^(1+ilog10(n))+n)/n,integer)

%p then a:=n; b:=b union {n}; print(n); break;

%p fi; fi; od; od; end: P(10^12);

%Y Cf. A171785, A249398, A250745, A250746, A250747.

%K nonn,base

%O 1,2

%A _Paolo P. Lava_, Dec 01 2014