A214437 - OEIS

%I #48 Jan 14 2021 20:36:20

%S 1,10,102,1020,10200,102000,1020005,10200056,102000564,1020005640,

%T 10200056405,102006162060,1020061620604,10200616206046,

%U 102006162060465,1020061620604656,10200616206046568,108054801036000018,1080548010360000180,10805480103600001800

%N Least numbers whose groups of 2,3,...,n digits taken from the left are divisible by 2,3,...,n.

%C The first 11 terms of the sequence are coincident with A078282.

%C a(6) is formed with 66,7 % zeros;  A(5) with 60 %; a(7) with 57,1 %; a(4), a(8), a(10) and a(20) with 50 %.

%C a(n) is the first term of A144688 with n digits, except that A144688 includes zero as first term. - _Franklin T. Adams-Watters_, Jul 18 2012

%C There are 25 terms in the sequence; the 25-digit number 3608528850368400786036725 is the last number to satisfy the requirements. - _Shyam Sunder Gupta_, Aug 04 2013

%H Shyam Sunder Gupta, <a href="/A214437/b214437.txt">Table of n, a(n) for n = 1..25</a>

%e a(6) = 102000 because 10, 102, 1020, 10200 and 102000 are divisible by 2, 3, 4, 5 and 6.

%e There are nine one-digit numbers that are divisible by 1; the smallest is 1, so a(1)=1.

%e For two-digit numbers, the second digit must be even, i.e., 0,2,4,6,8 to make it divisible by 2, which gives 10 as the smallest number to satisfy the requirement, so a(2)=10. - _Shyam Sunder Gupta_, Aug 04 2013

%t a=Table[j, {j, 9}]; r=2; t={};

%t While[!a == {}, n=Length[a]; nmin=Last[a]; k=1; b={};

%t While[!k>n, z0=a[[k]]; Do[z=10*z0+j; If[Mod[z, r]==0, b=Append[b, z]], {j, 0, 9}]; k++]; AppendTo[t, nmin]; a=b; r++]; t (* _Shyam Sunder Gupta_, Aug 04 2013 *)

%Y Cf. A078282, A158242, A144688.

%K nonn,base,fini,full

%O 1,2

%A _Robin Garcia_, Jul 17 2012