cp's OEIS Frontend

A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.

%I A052983 #17 Sep 04 2016 00:29:21
%S A052983 10,10,1110,100,10,1110,1111110,1000,1111111110,10,110,11100,1111110,
%T A052983 1111110,1110,10000,11111111111111110,1111111110,1111111111111111110,
%U A052983 100,1111110,110,11111111111111111111110,111000,100,1111110,1111111111111111111111111110
%N A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.
%C A052983 All entries are differences of two terms of A000042. Since the pigeonhole principle guarantees that, for any m, two among the first m+1 entries of A000042 are congruent modulo m, their difference (i.e. belonging to this sequence) is therefore divisible by m, so that such numbers exist for all m. This sequence is thus infinite.
%C A052983 For n>1, a(n) consists of s 1's and t 0's, where s=A084681(X) and t is the greater of p or q (s=1 for X=1, t=1 for p=q=0), when we write n=X*Y with (X,Y)=1 and Y=2^p*5^q.
%F A052983 a(n) = A276348(n) * n; A227362(a(n)) = 10. - _Jaroslav Krizek_, Aug 30 2016
%e A052983 We have a(6)=1110 because 6 divides 1110=6*185, the smallest such one with a string of 1's followed by that of 0's
%t A052983 f[n_] := Select[ Map[ FromDigits, IntegerDigits[ Table[ Sum[2^i, {i, k, j, -1}], {j, k, 1, -1}], 2]]/n, IntegerQ[ # ] & ]; g[n_] := Block[{k = 1}, While[ f[n] == {}, k++ ]; n*Min[ f[n]]]; Table[ g[n], {n, 1, 27}]
%t A052983 nn=30;With[{nos=Sort[Flatten[Table[FromDigits[Join[Table[1,{n}], Table[ 0,{i}]]],{n,nn},{i,5}]]]},Flatten[Table[Select[nos,Divisible[#,n]&,1],{n,nn}]]] (* _Harvey P. Dale_, Mar 09 2014 *)
%Y A052983 Cf. A000042, A002371, A007732.
%Y A052983 Cf. A084681.
%K A052983 base,nonn
%O A052983 1,1
%A A052983 _Lekraj Beedassy_, Jun 26 2003
%E A052983 Edited, corrected and extended by _Robert G. Wilson v_, Jun 26 2003