login
a(1)=1; thereafter, a(n) = smallest k such that the decimal concatenation [a(n-2)+1 a(n-2)+2, ... a(n-1)] divides the decimal concatenation [a(n-1)+1 a(n-1)+2 ... k].
1

%I #18 Dec 26 2024 03:46:08

%S 1,2,4,8,36

%N a(1)=1; thereafter, a(n) = smallest k such that the decimal concatenation [a(n-2)+1 a(n-2)+2, ... a(n-1)] divides the decimal concatenation [a(n-1)+1 a(n-1)+2 ... k].

%C a(6), if it exists, is > 10^6. - _Lars Blomberg_, Dec 01 2016

%C a(6) <= 86794654347484748866500883685475458354620023089553379437308257589024531796179370608623026912768. - _Max Alekseyev_, Dec 25 2024

%e n=3: a(3) = 4 because k=4 is the smallest number such that 2 divides the concatenation 345...k.

%e n=4: a(4) = 8 because k=8 is the smallest number such that 34 divides the concatenation 567...k. See A002782 for the relevant concatenations.

%Y Cf. A002782, A029455, A332580.

%K nonn,base,hard,more

%O 1,2

%A _N. J. A. Sloane_, Oct 23 2015