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A177023
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a(n) = 2^(2*n) mod (2*n+1).
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2
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1, 1, 1, 4, 1, 1, 4, 1, 1, 4, 1, 16, 13, 1, 1, 4, 9, 1, 4, 1, 1, 31, 1, 15, 4, 1, 49, 4, 1, 1, 4, 16, 1, 4, 1, 1, 34, 9, 1, 40, 1, 16, 4, 1, 64, 4, 54, 1, 58, 1, 1, 46, 1, 1, 4, 1, 39, 22, 30, 56, 4, 91, 1, 4, 1, 64, 94, 1, 1, 4, 114, 16, 25, 1, 1, 103, 109, 1, 4, 156, 1, 16, 1, 40, 85, 1, 134
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OFFSET
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1,4
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COMMENTS
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It is known that a(n) equals 1 when 2*n+1 is prime as a result of Fermat's little theorem. If not then a(n) equals 1 when 2*n+1 is a pseudoprime to base 2.
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LINKS
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FORMULA
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a(n) = 2^(2*n) mod (2*n+1) or a(n) = 4^n mod (2*n+1)
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EXAMPLE
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a(3) = 2^(2 * 3) mod (2 * 3 + 1) = 64 mod 7 = 1.
a(4) = 2^(2 * 4) mod (2 * 4 + 1) = 256 mod 9 = 4.
a(5) = 2^(2 * 5) mod (2 * 5 + 1) = 1024 mod 11 = 1.
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MAPLE
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MATHEMATICA
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Table[PowerMod[2, 2n, 2n + 1], {n, 90}] (* Harvey P. Dale, May 09 2012 *)
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PROG
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(Literate Haskell) > map (\k -> 2^(2*k) `mod` (2*k+1)) [1..100]
(PARI) a(n) = lift(Mod(4, 2*n+1)^n); \\ Michel Marcus, Jan 15 2018
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Nikolay Ulyanov (ulyanick(AT)gmail.com), May 01 2010
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STATUS
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approved
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