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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 }. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS For n > 8, a(n) = a(n-8) + 11111111, the first 8 values are in the data. 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 }. If k of the form 2*j+1, n*10^(2*j+1)-1 is divisible by 11. If k of the form 8*j, n*10^(8*j)-1 is divisible by 137. If k of the form 4*j+2, n*10^(4*j+2))-1 is divisible by 101. If k 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 }. LINKS Giovanni Resta, Table of n, a(n) for n = 1..27 FORMULA for n > 8, a(n) = a(n-8) + 11111111. CROSSREFS Cf. A243974, A244070. Sequence in context: A120409 A253466 A225811 * A083633 A043649 A282948 Adjacent sequences: A244345 A244346 A244347 * A244349 A244350 A244351 KEYWORD nonn,easy AUTHOR Pierre CAMI, Jun 28 2014 EXTENSIONS More terms from Giovanni Resta, Nov 23 2019 STATUS approved

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Last modified February 22 18:56 EST 2024. Contains 370260 sequences. (Running on oeis4.)