OFFSET
1,1
COMMENTS
For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are in the data.
If n is of the form 3*m+2, n*10^k+1 is always divisible by 3 but also has a divisor in the set { 11, 73, 101, 137 }.
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000 (a(1) .. a(27) from Giovanni Resta)
FORMULA
For n > 8, a(n) = a(n-8) + 11111111.
EXAMPLE
Consider n = 162207.
If k is of the form 2*j+1, n*10^(2*j+1)+1 is divisible by 11.
If k is of the form 8*j, n*10^(8*j)+1 is divisible by 137.
If k is of the form 4*j+2, n*10^(4*j+2)+1 is divisible by 101.
If k is of the form 8*j+4 then n*10^(8*j+4)+1 is divisible by 73.
This covers all k, so the covering set is { 11, 73, 101, 137 }.
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Pierre CAMI, Jun 28 2014
EXTENSIONS
More terms from Giovanni Resta, Nov 23 2019
STATUS
approved