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A320465
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a(n) = 2^n - (2^(n-1) mod n), where "mod" is the nonnegative remainder operator.
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1
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2, 4, 7, 16, 31, 62, 127, 256, 508, 1022, 2047, 4088, 8191, 16382, 32764, 65536, 131071, 262130, 524287, 1048568, 2097148, 4194302, 8388607, 16777208, 33554416, 67108862, 134217715, 268435448, 536870911, 1073741822, 2147483647, 4294967296, 8589934588
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OFFSET
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1,1
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COMMENTS
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Thomas Ordowski (private communication) observes that a(n) appears to be composite whenever n is composite. Is there any prime a(n) for composite n ?
Conjecture: for n > 2, a(n) is prime if and only if n is in A000043. Note that a(n) = 2^n - 1 if and only if n is an odd prime or pseudoprime to base 2 (A001567), so a counterexample cannot be a Fermat pseudoprime to base 2. - Thomas Ordowski, Oct 14 2018
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LINKS
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FORMULA
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For k >= 0, a(2^k) = 2^(2^k) = A001146(k). For n > 1, a(prime(n)) = 2^prime(n) - 1 = A001348(n). If p is an odd prime, then a(2p) = 4^p - 2. - Thomas Ordowski, Oct 14 2018
a(n) = 2^n - 1 = A000225(n) for n in A065091 U A001567 (odd prime or pseudoprime to base 2 = Sarrus or Poulet number).
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PROG
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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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