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A033714
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Number of zeros in numbers 0 to 999..9 (n digits).
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5
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1, 10, 190, 2890, 38890, 488890, 5888890, 68888890, 788888890, 8888888890, 98888888890, 1088888888890, 11888888888890, 128888888888890, 1388888888888890, 14888888888888890, 158888888888888890, 1688888888888888890, 17888888888888888890, 188888888888888888890
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OFFSET
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1,2
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COMMENTS
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This sequence also gives the total count of digits of n below 10^n. In such counts it makes sense to omit 10^0 as we are interested in having ten digits under each power of 10. For each power of 10 the total number of digits 0-9 is always the total of zeros for the next power. For example, at 10^1 there is 1 of each numeral 0-9, total 10 digits. At 10^2, the number of zeros is 10, with 20 each for the other 9 numerals and so on. - Enoch Haga, May 13 2006
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LINKS
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FORMULA
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a(n) = 10^(n-1)*n - (1/9)*10^n + 10/9. - Robert Israel, Jun 30 2014
G.f.: -x*(100*x^2-11*x+1) / ((x-1)*(10*x-1)^2). - Colin Barker, Jan 27 2015
a(n+1) = a(n) + 9 * A053541(n). (End)
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MATHEMATICA
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a[1] = 1; a[n_] := a[n] = 9*10^(n-2)*(n-1) + a[n-1]; Table[a[n], {n, 1, 17}] (* Jean-François Alcover, Jul 13 2012 *)
f[n_] := 1 + Sum[9 m*10^(m - 1), {m, n}]; Array[f, 18, 0] (* Robert G. Wilson v, Jun 29 2014 *)
LinearRecurrence[{21, -120, 100}, {1, 10, 190}, 20] (* Harvey P. Dale, Dec 03 2021 *)
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PROG
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(PARI) Vec(-x*(100*x^2-11*x+1)/((x-1)*(10*x-1)^2) + O(x^100)) \\ Colin Barker, Jan 27 2015
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CROSSREFS
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KEYWORD
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nonn,base,nice,easy
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AUTHOR
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Olivier Gorin (gorin(AT)roazhon.inra.fr)
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EXTENSIONS
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STATUS
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approved
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