A052038 - OEIS

%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