A243969
Integers n not of form 3m+2 such that for any integer k > 0, n*10^k+1 has a divisor in the set { 7, 11, 13, 37 }.
Original entry on oeis.org
9175, 9351, 17676, 24826, 26038, 28612, 38026, 38158, 46212, 46927, 48247, 56473, 61863, 63075, 63898, 65649, 75063, 75195, 83425, 83964, 85284, 91750, 93510, 100935
Offset: 1
9175*10^k+1 is divisible by 11 for k of form 6m+1, 6m+3, 6m+5, by 37 for k of form 6m (and also 6m+3), by 13 for 6m+2, and by 7 for 6m+4. This covers all k. {7, 11, 13, 37} is called a covering set. - _Jens Kruse Andersen_, Jul 09 2014
A244348
Integers n such that for every integer k>0, n*10^k+1 has a divisor in the set { 11, 73, 101, 137 }.
Original entry on oeis.org
162207, 1622070, 3349554, 5109589, 6651446, 7001622, 9589051, 10958905, 11273318, 12733181, 14460665, 16220700, 17762557, 18112733, 20700162, 22070016, 22384429, 23844292, 25571776, 27331811, 28873668, 29223844, 31811273, 33181127, 33495540, 34955403, 36682887
Offset: 1
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 }.
A244598
Integers n such that for every k > 0, n*10^k-1 has a divisor in the set { 11, 73, 101, 137 }.
Original entry on oeis.org
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
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 }.
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