login
A244549
Integers n such that for every integer k>0, n*6^k+1 has a divisor in the set { 7, 13, 31, 37, 97 }.
0
174308, 188299, 702703, 1045848, 1129794, 1615907, 1956746, 2485141, 3162650, 4216218, 4786277, 4800566, 5048170, 6275088, 6778764, 7075837, 7276821, 7549807, 8468524, 8554258, 8851331, 9616447, 9695442, 10039882
OFFSET
1,1
COMMENTS
For n > 24 a(n) = a(n-24) + 10124569, 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 have always a divisor in the set { 7, 13, 31, 37, 97 } for every k.
FORMULA
For n > 24 a(n) = a(n-24) + 10124569.
KEYWORD
nonn
AUTHOR
Pierre CAMI, Jun 29 2014
STATUS
approved