|
|
A277057
|
|
Least k such that n-th repunit times k contains all digits from 1 to 9.
|
|
3
|
|
|
123456789, 11225079, 1113198, 210789, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115, 11115
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Starting from n=5, a(n)=A277059(10)=11115, and the corresponding pandigital numbers are 123499...98765 (n-4 nines). Actually, for all n, the resulting numbers are zeroless pandigitals.
a(1)-a(5) constitute row 10 of A277058.
a(n)*A002275(n) is 123456789, 123475869, 123564978, 234186579, 123498765, ...
|
|
LINKS
|
|
|
EXAMPLE
|
a(2) = 11225079 because A002275(2)*11225079 = 11*11225079 = 123475869 that contains all digits from 1 to 9 and 11225079 is the least number with this property.
|
|
PROG
|
(PARI) isok(n) = my(d=digits(n)); vecmin(d) && (#Set(digits(n)) == 9);
a(n) = {if (n==1, return(123456789)); my(k=1); while(! isok(k*(10^n - 1)/9), k++); k; } \\ Michel Marcus, Sep 26 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|