login
A244598
Integers n such that for every k > 0, n*10^k-1 has a divisor in the set { 11, 73, 101, 137 }.
0
152206, 1522060, 4109489, 4459665, 6001522, 7761557, 9489041, 10948904, 11263317, 12633171, 15220600, 15570776, 17112633, 18872668, 20600152, 22060015, 22374428, 23744282, 26331711, 26681887, 28223744, 29983779, 31711263, 33171126, 33485539, 34855393, 37442822
OFFSET
1,1
COMMENTS
For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are given in the data.
If n is of the form 3*m+1 then n*10^k-1 is always divisible by 3 but also has a divisor in the set { 11, 73, 101, 137 }.
FORMULA
For n > 8, a(n) = a(n-8) + 11111111.
EXAMPLE
Consider n = 152206.
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 73.
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, n*10^(8*j+4)-1 is divisible by 137.
This covers all k, so the covering set is { 11, 73, 101, 137 }.
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Pierre CAMI, Jul 01 2014
EXTENSIONS
a(9)-a(27) from Jason Yuen, Nov 10 2024
STATUS
approved