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 A214437 Least numbers whose groups of 2,3,...,n digits taken from the left are divisible by 2,3,...,n. 4
 1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640, 10200056405, 102006162060, 1020061620604, 10200616206046, 102006162060465, 1020061620604656, 10200616206046568, 108054801036000018, 1080548010360000180, 10805480103600001800 (list; graph; refs; listen; history; text; internal format)
 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,changed AUTHOR Robin Garcia, Jul 17 2012 STATUS approved

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Last modified January 25 12:00 EST 2021. Contains 340416 sequences. (Running on oeis4.)