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A244348 Integers n such that for every integer k>0, n*10^k+1 has a divisor in the set { 11, 73, 101, 137 }.

%I A244348 #59 Nov 10 2024 09:25:37
%S A244348 162207,1622070,3349554,5109589,6651446,7001622,9589051,10958905,
%T A244348 11273318,12733181,14460665,16220700,17762557,18112733,20700162,
%U A244348 22070016,22384429,23844292,25571776,27331811,28873668,29223844,31811273,33181127,33495540,34955403,36682887
%N A244348 Integers n such that for every integer k>0, n*10^k+1 has a divisor in the set { 11, 73, 101, 137 }.
%C A244348 For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are in the data.
%C A244348 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 }.
%H A244348 Robert Israel, <a href="/A244348/b244348.txt">Table of n, a(n) for n = 1..10000</a> (a(1) .. a(27) from _Giovanni Resta_)
%F A244348 For n > 8, a(n) = a(n-8) + 11111111.
%e A244348 Consider n = 162207.
%e A244348 If k is of the form 2*j+1, n*10^(2*j+1)+1 is divisible by 11.
%e A244348 If k is of the form 8*j, n*10^(8*j)+1 is divisible by 137.
%e A244348 If k is of the form 4*j+2, n*10^(4*j+2)+1 is divisible by 101.
%e A244348 If k is of the form 8*j+4 then n*10^(8*j+4)+1 is divisible by 73.
%e A244348 This covers all k, so the covering set is { 11, 73, 101, 137 }.
%Y A244348 Cf. A243974, A244070.
%K A244348 nonn,easy
%O A244348 1,1
%A A244348 _Pierre CAMI_, Jun 28 2014
%E A244348 More terms from _Giovanni Resta_, Nov 23 2019