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A177916
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Numbers k such that k^3 divides 16^(k^2) - 1.
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16
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1, 3, 5, 15, 21, 39, 55, 57, 105, 111, 155, 165, 195, 205, 219, 273, 285, 327, 399, 465, 505, 555, 609, 615, 741, 777, 903, 915, 1095, 1155, 1255, 1265, 1365, 1443, 1515, 1533, 1635, 1705, 1995, 2067, 2109, 2145, 2255, 2265, 2289, 2373, 2667, 2715, 2847
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OFFSET
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1,2
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COMMENTS
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The only primes in the sequence are 3 and 5.
If m and n are in the sequence and are coprime, then m*n is in the sequence.
Are all members of the sequence divisible by 3 or 5? (End)
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LINKS
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MAPLE
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MATHEMATICA
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Select[Range[3000], IntegerQ[(16^(#^2) - 1) / #^3 ] &] (* Vincenzo Librandi, Apr 24 2015 *)
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PROG
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(Magma) [n: n in [1..2000] | Denominator((16^(n^2) - 1)/n^3) eq 1]; // Vincenzo Librandi, Apr 24 2015
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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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