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A198971
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a(n) = 5*10^n - 1.
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8
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4, 49, 499, 4999, 49999, 499999, 4999999, 49999999, 499999999, 4999999999, 49999999999, 499999999999, 4999999999999, 49999999999999, 499999999999999, 4999999999999999, 49999999999999999, 499999999999999999, 4999999999999999999
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OFFSET
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0,1
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COMMENTS
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Also maximal value of GCD of 2 distinct (n+1)-digit numbers (compare with A126687). - Michel Marcus, Jun 24 2013
Also, a(n) is the largest obtained remainder when an (n+1)-digit number m is divided by any k with 1 <= k <= m. This remainder is obtained when 10^(n+1)-1 is divided by 5*10^n, example: 999 = 500 * 1 + 499, and a(2) = 499. - Bernard Schott, Nov 23 2021
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LINKS
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FORMULA
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a(n) = 10*a(n-1) + 9.
a(n) = 11*a(n-1) - 10*a(n-2), n>1.
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MATHEMATICA
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CoefficientList[Series[(4 + 5*x)/(1 - 11*x + 10*x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 03 2013 *)
LinearRecurrence[{11, -10}, {4, 49}, 20] (* Harvey P. Dale, Dec 30 2018 *)
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PROG
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(Magma) [5*10^n-1 : n in [0..20]]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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