%I #15 Dec 07 2019 03:26:30
%S 1,5,3,2,2,1,1,1,1,1,9,8,7,7,6,6,5,5,5,5,4,4,4,4,4,3,3,3,3,3,3,3,3,2,
%T 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,9,9,9,9,9
%N First nonzero digit in expansion of 1/n.
%C The number of times each digit occurs for numbers < 10^k:
%C ...\a(n)==1.........2.......3........4........5........6........7........8........9
%C 10^k\
%C 1.........5.........2........1........0........1........0........0........0........0
%C 2........55........19........9........5........5........2........2........1........1
%C 3.......555.......186.......92.......55.......39.......26.......19.......15.......12
%C 4......5555......1853......925......555......373......264......197......154......123
%C 5.....55555.....18520.....9258.....5555.....3707.....2645.....1982.....1543.....1234
%C 6....555556....185187....92591....55555....37041....26454....19839....15432....12345
%C 7...5555555...1851854...925924...555555...370375...264549...198410...154321...123456
%C 8..55555555..18518521..9259257..5555555..3703709..2645501..1984124..1543210..1234567
%C 9.555555555.185185188.92592590.55555555.37037043.26455025.19841266.15432099.12345678
%C ...
%C Inf. ...5/9......5/27.....5/54.....5/90.....1/27........?........?........?........?
%F a(n) = floor(10^floor(1+log_10(n-1))/n). After 10^k terms the number of times m will have appeared will be about 10^(k+1)/(9*m*(m+1)), e.g., 1 will appear just over 55.5% of the time. - _Henry Bottomley_, May 11 2001
%F a(n) = A000030(floor(A011557(k)/n)) for k >= A004218(n). - _Reinhard Zumkeller_, Feb 27 2011
%t f[n_] := RealDigits[1/n, 10, 12][[1, 1]]; Array[f, 105]
%Y Cf. A052039, A033330, A033420, A061861.
%K nonn,base,easy
%O 1,2
%A _Patrick De Geest_, Dec 15 1999