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A052379
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Number of integers from 1 to 10^(n+1)-1 that lack 0 and 1 as a digit.
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11
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8, 72, 584, 4680, 37448, 299592, 2396744, 19173960, 153391688, 1227133512, 9817068104, 78536544840, 628292358728, 5026338869832, 40210710958664, 321685687669320, 2573485501354568, 20587884010836552, 164703072086692424, 1317624576693539400, 10540996613548315208
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OFFSET
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0,1
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LINKS
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FORMULA
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a(n) = (8^(n+2) - 1)/7 - 1.
a(n) = 8*a(n-1) + 8.
a(n) = Sum_{k=1..n} 8^k. - corrected by Michel Marcus, Sep 25 2014
Conjecture: a(n) = A023001(n+2)-1. - R. J. Mathar, May 18 2007. Comment from Vim Wenders, Mar 26 2008: The conjecture is true: the g.f. leads to the closed form a(n) = -(8/7)*(1^n) + (64/7)*(8^n) = (-8 + 64*8^n)/7 = (-8 + 8^(n+2))/7 = (8^(n+2) - 1)/7 - 1 = A023001(n+2) - 1.
a(n) = 9*a(n-1) - 8*a(n-2); a(0)=8, a(1)=72. - Harvey P. Dale, Sep 22 2013
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EXAMPLE
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For n=1, the numbers from 1 to 99 which have 0 or 1 as a digit are the numbers 1 and 10, 20, 30, ..., 90 and 11, 12, ..., 18, 19 and 21, 31, ..., 91. So a(1) = 99 - 27 = 72.
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MAPLE
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MATHEMATICA
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(8^(Range[0, 20]+2)-1)/7-1 (* or *) LinearRecurrence[{9, -8}, {8, 72}, 20] (* Harvey P. Dale, Sep 22 2013 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,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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