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

Author

Robin Garcia, Jul 17 2012

Keywords

• nonn
• base
• fini
• full

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

Examples

			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
		

Links

• Shyam Sunder Gupta, Table of n, a(n) for n = 1..25
• Shyam Sunder Gupta, On Some Special Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 22, 527-565.

Crossrefs

Cf. A078282, A158242, A144688.

Programs

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 *)