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Search: id:A109012
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| 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Start with positive integer n. At each step, either (a) multiply by any positive integer or (b) remove all zeros from the number. a(n) is the smallest number that can be reached by this process. - David W. Wilson (davidwwilson(AT)comcast.net), Nov 01 2005
Comment from Martin Fuller (martin_n_fuller(AT)btinternet.com), Jul 09, 2007: (Start) Also the minimal positive difference between numbers whose digit sum is a multiple of n. Proof:
Construction: Pick a positive number which does not finish with 9,
and has a digit sum n-a(n). To form the lower number, append 9
until the digit sum is a multiple of n. This is always possible
since the difference is gcd(n,9). Add a(n) to form the higher
number, which will have digit sum n.
E.g. n=12: prefix=18, lower=18999, higher=19002, difference=3.
Minimality: All numbers are a multiple of a(n) if their digit sum
is a multiple of n. Hence the minimal difference is at least a(n). (End)
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FORMULA
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a(n) = 1 + 2*[3|n] + 6*[9|n], where [x|y] = 1 when x divides y, 0 otherwise.
a(n) = a(n-9).
Multiplicative with a(p^e, 9) = GCD(p^e, 9). David W. Wilson (davidwwilson(AT)comcast.net) Jun 12, 2005.
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CROSSREFS
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Cf. A109004.
Sequence in context: A006084 A059928 A010163 this_sequence A037478 A010162 A072224
Adjacent sequences: A109009 A109010 A109011 this_sequence A109013 A109014 A109015
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KEYWORD
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nonn,easy,mult
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AUTHOR
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Mitch Harris (Harris.Mitchell(AT)mgh.harvard.edu)
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