OFFSET
1,2
COMMENTS
The number of times each digit occurs for numbers < 10^k:
...\a(n)==1.........2.......3........4........5........6........7........8........9
10^k\
1.........5.........2........1........0........1........0........0........0........0
2........55........19........9........5........5........2........2........1........1
3.......555.......186.......92.......55.......39.......26.......19.......15.......12
4......5555......1853......925......555......373......264......197......154......123
5.....55555.....18520.....9258.....5555.....3707.....2645.....1982.....1543.....1234
6....555556....185187....92591....55555....37041....26454....19839....15432....12345
7...5555555...1851854...925924...555555...370375...264549...198410...154321...123456
8..55555555..18518521..9259257..5555555..3703709..2645501..1984124..1543210..1234567
9.555555555.185185188.92592590.55555555.37037043.26455025.19841266.15432099.12345678
...
Inf. ...5/9......5/27.....5/54.....5/90.....1/27........?........?........?........?
FORMULA
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
MATHEMATICA
f[n_] := RealDigits[1/n, 10, 12][[1, 1]]; Array[f, 105]
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Patrick De Geest, Dec 15 1999
STATUS
approved