login
A244545
Integers n such that for every integer k>0, n*6^k+1 has a divisor in the set { 7, 13, 31, 37, 43 }.
1
243417, 1161910, 1293662, 1434861, 1446213, 1460502, 1473746, 1689722, 2284675, 2483249, 2485141, 2693347, 2695449, 2708061, 2783733, 3207751, 3237765, 3273761, 3684535, 4120955, 4154366, 4189067, 4274801, 4354265
OFFSET
1,1
COMMENTS
For n > 24, a(n) = a(n-24) + 4488211, the first 24 values are in the data.
When the number a(n) has 4 or 9 as the last digit, the number a(n)*6^k-1 is always divisible by 5 and always has a divisor in the set { 7, 13, 31, 37, 97 } for every k.
FORMULA
For n > 24 a(n) = a(n-24) + 4488211.
KEYWORD
nonn
AUTHOR
Pierre CAMI, Jun 29 2014
STATUS
approved