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A235400
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a(1)=1. For n>1, assume a(n-1) has decimal expansion qrstuvwxyz (with at most ten digits, where some of q, r, s, ... may be zero). Then a(n) = sum 99...9 (with z 9's) + 88...8 (with y 8's) + 77...7 (with x 7's) + ... For example, if a(n-1) were 243, we would sum 77 + 8888 + 999 and get 9964.
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3
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1, 9, 999999999, 4999999995, 4000099995, 2888988885, 800099995, 2900099996, 3000999996, 3334333329, 1000006195, 889655560, 368199996, 2923444555, 111394795, 1452329335, 666792195, 1449655625, 334781195, 948992565, 1145513395, 889070335, 372323885, 168990195
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OFFSET
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1,2
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COMMENTS
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The sequence is periodic: a(470) = a(312) = 8005, etc.
See A235460 for the analogous sequence where we number the digits starting with 1 at the leftmost digit.
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REFERENCES
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Rodolfo Kurchan, Problems with repdigits, to appear in Proceedings of Gathering for Gardner 2014 conference
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LINKS
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EXAMPLE
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1 leads to one 9, then 9 leads to 999999999, then we have to sum 111111111 + 222222222 + ... + 999999999 = 4999999995, then we have to sum 1111 + 222222222 + ... + 888888888 + 99999 = 4000099995, and so on.
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MATHEMATICA
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nxt[n_]:=Total[FromDigits/@(Table[#[[1]], {#[[2]]}]&/@Thread[{ Range[ 9, 10-IntegerLength[n], -1], Reverse[IntegerDigits[n]]}])]; NestList[ nxt, 1, 30] (* Harvey P. Dale, Apr 24 2015 *)
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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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STATUS
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approved
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