A249067 Smallest k such that {first digit of j*n, 0<=j<=k} = {0,1,...,9}.

9, 45, 27, 23, 18, 15, 13, 12, 9, 9, 9, 42, 62, 43, 54, 44, 53, 45, 43, 45, 43, 41, 35, 34, 36, 35, 34, 33, 32, 27, 26, 25, 25, 27, 26, 25, 25, 24, 24, 23, 22, 22, 21, 21, 18, 18, 18, 17, 17, 18, 18, 18, 17, 17, 17, 17, 16, 16, 16, 15, 15, 15, 15, 15, 14, 14, 14, 14, 14, 13, 13, 13, 13, 13, 12, 12, 12, 12, 12, 12, 12, 11, 11, 11, 11, 11, 11, 11

Offset

1,1

Comments

In other words: a(n) = how long until you have seen every first-digit that you can in the n times table.

Conjecture: a(n) <= N for all n. Perhaps N can be taken as 81. - Charles R Greathouse IV, Oct 20 2014

The conjecture above is true. In fact, a(n) takes on precisely 37 values: 9, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 51, 52, 53, 54, 61, 62, 63, 71, 72, and 81. The last of these to occur is a(291) = 31. - Charles R Greathouse IV, Mar 17 2018

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Example

a(2) = 45, because the first time a multiple of 2 starts with a 9 is 45*2 = 90, and all other first-digits occur earlier than that.

Maple

a:= proc(n) local j, s; s:={};
      for j while s<>{$1..9} do
        s:= s union {(p-> iquo(p, 10^(length(p)-1)))(j*n)}
      od; j-1
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Oct 20 2014

Prog

(Python)
def a(n):
...got = [0]*10
...i = 1
...while sum(got)<9:
......d=int(str(n*i)[0])
......got[d] = 1
......i += 1
...return i-1
(PARI) a(n)=my(v=vector(9), s, k, t); while(s<9, t=digits(k++*n)[1]; if(v[t]==0, v[t]=1; s++)); k \\ Charles R Greathouse IV, Oct 20 2014

Crossrefs

Sequence in context: A225488 A096688 A181431 * A366420 A124983 A087969

Adjacent sequences: A249064 A249065 A249066 * A249068 A249069 A249070

Keyword

nonn,base,look,easy

Author

Christian Perfect, Oct 20 2014

Status

approved