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%I #30 Apr 16 2018 03:21:22
%S 1,2,3,41,43,129,9567001,21147541,22662659,23817877,24837187,28850377,
%T 28872229,37916473,48749751,70416307,439229167,834385607,2270365163,
%U 2278377431,3751789547,4433933101,4810754611,14432263833,15632412757,30530543651,42441819717,65591903199,65857498407
%N a(1) = 1; a(2) = 2; then a(n) is the smallest number > a(n-1) such that a(n) divides concat(a(1), a(2), ..., a(n-1)).
%e a(3) = 3, which makes the concatenation of the first three terms: 123. After 3, the next-highest factor of 123 is 41, so a(4) = 41. The concatenation of the first four terms is then 12341. After 41, the next-highest factor of 12341 is 43, so a(5) = 43.
%p A[1]:= 1: A[2]:= 2: C:= 1:
%p for n from 3 to 20 do
%p C:= A[n-1]+C*10^(ilog10(A[n-1])+1);
%p A[n]:= min(select(`>`,numtheory:-divisors(C),A[n-1]))
%p od:
%p seq(A[i],i=1..20); # _Robert Israel_, Apr 12 2018
%Y Compare A240588, in which each term does not need to strictly increase as long as it has not yet appeared in the sequence.
%Y Compare also A171785, in which each term must divide the concatenation of all terms in the sequence including itself.
%Y In A029455, each term divides the concatenation of all smaller positive integers.
%Y In A110740, each term divides the concatenation of all strictly smaller positive integers.
%K nonn,base
%O 1,2
%A _Daniel Sterman_, Apr 11 2018
%E a(16)-a(20) from _Robert Israel_, Apr 12 2018
%E a(21)-a(29) from _Daniel Suteu_, Apr 12 2018
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