A214437 Least numbers whose groups of 2,3,...,n digits taken from the left are divisible by 2,3,...,n.

1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800

Offset

1,2

Comments

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

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 %.

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

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

Links

Shyam Sunder Gupta, Table of n, a(n) for n = 1..25

Example

a(6) = 102000 because 10, 102, 1020, 10200 and 102000 are divisible by 2, 3, 4, 5 and 6.
There are nine one-digit numbers that are divisible by 1; the smallest is 1, so a(1)=1.
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

Mathematica

a=Table[j, {j, 9}]; r=2; t={};
While[!a == {}, n=Length[a]; nmin=Last[a]; k=1; b={};
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 *)

Crossrefs

Cf. A078282, A158242, A144688.

Sequence in context: A080502 A078283 A045874 * A078282 A336401 A037503

Adjacent sequences: A214434 A214435 A214436 * A214438 A214439 A214440

Keyword

nonn,base,fini,full

Author

Robin Garcia, Jul 17 2012

Status

approved