

A214437


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


2



1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800
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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. AdamsWatters, Jul 18 2012
There are 25 terms in the sequence; the 25digit 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 onedigit numbers divisible by 1 and smallest is 1 so a(1)=1,For 2 digit numbers, second digit must be even i.e.0,2,4,6,8 that make it divisible by 2,which gives 10 as smallest 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 A037503 A037680
Adjacent sequences: A214434 A214435 A214436 * A214438 A214439 A214440


KEYWORD

nonn,base,fini,full


AUTHOR

Robin Garcia, Jul 17 2012


STATUS

approved



