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A195470
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Number of numbers k with 0 <= k < n such that 2^k + 1 is multiple of n.
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4
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1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 2, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 1, 0, 1, 0, 0, 0, 5, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0
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OFFSET
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1,17
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COMMENTS
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LINKS
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EXAMPLE
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a(1)=#{0}=1, (2^0 + 1) mod 1;
a(17) = #{4, 12} = 2, (2^4 + 1) mod 17 = (2^12 + 1) mod 17 = 0;
a(18) = #{} = 0;
a(19) = #{9} = 1, (2^9 + 1) mod 19 = 0.
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MATHEMATICA
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nn = 100; pwrs = 2^Range[0, nn] + 1; Table[cnt = 0; Do[If[Mod[pwrs[[i]], n] == 0, cnt++], {i, n}]; cnt, {n, nn}] (* T. D. Noe, Sep 30 2011 *)
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PROG
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(Haskell)
a195470 n = length $ filter ((== 0) . (`mod` n)) $
take (fromInteger n) a000051_list
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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