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A069056
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Numbers k such that Sum_{d|k} d^2*mu(d) divides k^2.
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1
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1, 12, 24, 36, 48, 72, 96, 108, 120, 144, 192, 216, 240, 288, 324, 336, 360, 384, 432, 480, 576, 600, 648, 672, 720, 768, 864, 960, 972, 1008, 1080, 1152, 1200, 1296, 1344, 1440, 1536, 1728, 1800, 1920, 1944, 2016, 2160, 2304, 2352, 2400, 2448, 2592, 2688
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OFFSET
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1,2
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COMMENTS
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If n > 1, a(n+1) - a(n) == 0 (mod 12), so a(n+1) - a(n) = 12 for n=2,3,4,5,7,8,...; a(n+1) - a(n) = 24 for n=6,9,.... Conjecture: if c > 2 and n > 1, Sum_{d|n} d^c*mu(d) never divides n^c. Hence A063453(n) never divides n^3 for n > 1.
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LINKS
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FORMULA
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Numbers k such that A046970(k) divides k.
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MATHEMATICA
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f[d_] := d^2*MoebiusMu[d]; ok[n_] := Divisible[n^2, Total[f /@ Divisors[n]]]; Select[Range[3000], ok] (* Jean-François Alcover, Nov 15 2011 *)
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PROG
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(PARI) for(n=1, 1000, if(n^2%sumdiv(n, d, moebius(d)*d^2)==0, print1(n, ", ")))
(Haskell)
a069056 n = a069056_list !! (n-1)
a069056_list = filter (\x -> x ^ 2 `mod` a046970 x == 0) [1..]
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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