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A244471
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Lexicographically earliest sequence of integers with property that if a vertical line is drawn between any pair of adjacent digits, the number Z formed by the digits to the left of the line is divisible by the digit to the right of the line.
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3
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1, 11, 3, 7, 71, 31, 111, 113, 33, 117, 77, 13, 37, 711, 1111, 19, 9, 91, 1117, 73, 311, 131, 1131, 1133, 93, 331, 11111, 39, 99, 97, 119, 333, 911, 133, 931, 1139, 771, 337, 713, 339, 933, 391, 1137, 773, 1113, 991, 11171, 3111, 777, 3311, 79, 17, 191, 171, 11311, 137, 719, 993
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OFFSET
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1,2
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COMMENTS
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"Lexicographically earliest" means in the sense of a sequence of integers, not digits.
No digit can be even or five. - Hans Havermann, Jul 02 2014 [Proof: if not, let d be the first digit in the sequence that is even or 5, and let Z be the concatenation of all earlier digits. But then Z is odd and does not end in 5, so is not divisible by d. Contradiction. - N. J. A. Sloane, Jul 03 2014] So any term must have only the odd digits {1, 3, 7, 9} (see A136333). - Robert G. Wilson v, Jul 02 2014
We choose the next term, a(n), to be the minimal number not already in the sequence such that the property "if a vertical line is drawn between any pair of adjacent digits, the number Z formed by the digits to the left of the line is divisible by the first digit following Z" holds.
So even if Z is prime, the next term can start with a 1.
So if Z is divisible by any d in {2,3,...,9} the next term can start with 1 or d, otherwise it must start with 1.
This sequence is missing A136333 terms 313, 319, 373, 379, 717, 737, 797, 913, 919, 939, 973, 979, 1313, ... The earliest occurrences of n-digit numbers are the repunits at indices 1, 2, 7, 15, 27, 97, 372, 939, 2164, 4781, 10851, 22779, 47056, ... The latest n-digit numbers and their indices are: (9,17), (17,52), (397,290), (1917,867), (19317,2003), (199117,7241), (1999117,17953), (19999997,44173), ... - Hans Havermann, Jul 04 2014, Jul 07 2014, Jul 15 2014
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REFERENCES
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Eric Angelini, Posting to Sequence Fans Mailing List, Jun 26 2014
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LINKS
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EXAMPLE
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After 1,11,3,7, let a(5) = x be the next term. Now 11137 = 7*37*43, so x must begin with 1 or 7. The candidates for x are therefore 12,13,...,19,71,72,,...,79,111,...
If x=12, we would get 1 11 3 7 12 ... but Z = 11371 is prime and is not divisible by 2, ..., 9. So x is not 12, ...,19. The next candidate is x=71, and this works. So a(5)=71.
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MATHEMATICA
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r=f=e={1, 3, 7, 9}; Do[e=10*e; f=Flatten[Table[e[[i]]+f, {i, 4}]]; r=Join[r, f], {9}]; r=Select[r, Intersection[Partition[IntegerDigits[#], 3, 1], IntegerDigits[{313, 319, 373, 379, 717, 737, 797, 913, 919, 939, 973, 979}]]=={}&]; t=0; Do[c=1; While[d=IntegerDigits[r[[c]]]; Union[Table[IntegerQ[(10^i*t+FromDigits[Take[d, i]])/d[[i+1]]], {i, 0, Length[d]-1}]]!={True}, c++]; Print[r[[c]]]; t=10^Length[d]*t+r[[c]]; r=Delete[r, c], {10850}] (* Hans Havermann, Jul 04 2014 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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