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Integers n such that for every k > 0, n*10^k-1 has a divisor in the set { 11, 73, 101, 137 }.
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%I #27 Nov 10 2024 02:23:00

%S 152206,1522060,4109489,4459665,6001522,7761557,9489041,10948904,

%T 11263317,12633171,15220600,15570776,17112633,18872668,20600152,

%U 22060015,22374428,23744282,26331711,26681887,28223744,29983779,31711263,33171126,33485539,34855393,37442822

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

%C For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are given in the data.

%C 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 }.

%F For n > 8, a(n) = a(n-8) + 11111111.

%e Consider n = 152206.

%e If k is of the form 2*j+1, n*10^(2*j+1)-1 is divisible by 11.

%e If k is of the form 8*j, n*10^(8*j)-1 is divisible by 73.

%e If k is of the form 4*j+2, n*10^(4*j+2)-1 is divisible by 101.

%e If k is of the form 8*j+4, n*10^(8*j+4)-1 is divisible by 137.

%e This covers all k, so the covering set is { 11, 73, 101, 137 }.

%Y Cf. A243969, A243974, A244348.

%K nonn,easy

%O 1,1

%A _Pierre CAMI_, Jul 01 2014

%E a(9)-a(27) from _Jason Yuen_, Nov 10 2024