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A004290
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Least positive multiple of n that when written in base 10 uses only 0's and 1's.
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45
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1, 10, 111, 100, 10, 1110, 1001, 1000, 111111111, 10, 11, 11100, 1001, 10010, 1110, 10000, 11101, 1111111110, 11001, 100, 10101, 110, 110101, 111000, 100, 10010, 1101111111, 100100, 1101101, 1110, 111011, 100000, 111111, 111010
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OFFSET
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1,2
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COMMENTS
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It is easy to show that a(n) always exists and in fact has at most n digits [Wu, 2014]. - N. J. A. Sloane, Jun 13 2014
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LINKS
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FORMULA
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MAPLE
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f:= proc(n)
local L, x, m, r, k, j;
for x from 2 to n-1 do L[0, x]:= 0 od:
L[0, 0]:= 1: L[0, 1]:= 1;
for m from 1 do
if L[m-1, (-10^m) mod n] = 1 then break fi;
L[m, 0]:= 1;
for k from 1 to n-1 do
L[m, k]:= max(L[m-1, k], L[m-1, k-10^m mod n])
od;
od;
r:= 10^m; k:= -10^m mod n;
for j from m-1 by -1 to 1 do
if L[j-1, k] = 0 then
r:= r + 10^j; k:= k - 10^j mod n;
fi
od;
if k = 1 then r:= r + 1 fi;
r
end proc:
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MATHEMATICA
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a[n_] := For[k = 1, True, k++, b = FromDigits[ IntegerDigits[k, 2] ]; If[Mod[b, n] == 0, Return[b]]]; a[0] = 0; Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Jun 14 2013, after Reinhard Zumkeller *)
With[{c=Rest[Union[FromDigits/@Flatten[Table[Tuples[{1, 0}, i], {i, 10}], 1]]]}, Join[{0}, Flatten[ Table[ Select[c, Divisible[#, n]&, 1], {n, 40}]]]] (* Harvey P. Dale, Dec 07 2013 *)
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PROG
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(Haskell)
a004290 0 = 0
a004290 n = head [x | x <- tail a007088_list, mod x n == 0]
if n > 0:
for i in range(1, 2**n):
x = int(bin(i)[2:])
if not x % n:
return x
return 0
(PARI) a(n) = {if( n==0, return (0)); my(m = n); while (vecmax(digits(m)) != 1, m+=n); m; } \\ Michel Marcus, Feb 09 2016, May 27 2020
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CROSSREFS
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KEYWORD
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nonn,base,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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