A242811 - OEIS

%I #11 Sep 18 2017 02:41:15

%S 9,99,996,9968,99682,996822,9968228,36894671,368946712

%N a(n) is the largest n-digit number whose truncation after its first k digits is divisible by the k-th Lucas number (A000032(n)) for k = 1..n.

%C There are 9 terms in the series and 9-digit number 368946712 is the last number to satisfy the requirements.

%e   368946712 is divisible by Lucas(9) = 76;

%e    36894671 is divisible by Lucas(8) = 47;

%e     3689467 is divisible by Lucas(7) = 29;

%e      368946 is divisible by Lucas(6) = 18;

%e       36894 is divisible by Lucas(5) = 11;

%e        3689 is divisible by Lucas(4) =  7;

%e         368 is divisible by Lucas(3) =  4;

%e          36 is divisible by Lucas(2) =  3;

%e           3 is divisible by Lucas(1) =  1.

%t a=Table[j, {j, 3, 10, 2}]; r=2; t={}; While[!a == {}, n=Length[a]; nmax=Last[a]; k=1; b={}; While[!k>n, z0=a[[k]]; Do[z=10*z0+j; If[Mod[z, LucasL[r]]==0, b=Append[b, z]], {j, 0, 9}]; k++]; AppendTo[t, nmax]; a=b; r++]; t

%Y Cf. A000032, A225614, A242808, A242809, A242810.

%K nonn,base,fini,full

%O 1,1

%A _Michel Lagneau_, May 23 2014