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A353214
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a(n) = 2^A007013(4) mod prime(n); the last term of this sequences is when a(n) = 1.
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1
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0, -1, -2, 2, -4, -2, -8, 2, -5, -2, 4, -2, 5, 2, -11, -20, -22, 6, -23, -21, 2, -3, -16, -25, -31, 40, 19, -29, -2, -2, 2, -49, 19, 68, -56, -23, -59, 45, 29, -2, 62, 63, 27, 54, -2, -22, -46, 28, -85, -2, -29, 17, -113, -4, -128, -65, -46, 20, -51, -98, -64
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OFFSET
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1,3
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COMMENTS
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This sequence uses the centered version of mod. The residue system modulo prime(n) is {-1*floor(prime(n)/2)..floor(prime(n)/2)}. This is so that this sequence will encode information about the numbers around 2^A007013(4). If a(n) = k and prime(n) < 2^A007013(4) - k, then 2^A007013(4) - k is not prime (prime(n) is a factor of 2^A007013(4) - k). For example, a(22) = -3, so prime(22) = 79 is a factor of 2^A007013(4) + 3.
The length of this sequence is the lowest value of n such that A014664(n) = A007013(4). This is because for any power of 2, 2^p, if p == 0 (mod A014664(n)), then 2^p == 1 (mod prime(n)) (prime(n) is a factor of A000225(p)). Since A007013(4) is prime, we can apply this to get: If A014664(n) = A007013(4) and prime(n) < A007013(5), then A007013(5) is not prime (prime(n) is a nontrivial factor).
For any n such that prime(n) < 5*(10^51 + 5*10^9), a(n) != 1.
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LINKS
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FORMULA
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a(n) = 2^(2^127 - 1) mod prime(n).
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PROG
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(PARI) A353214(n)=my(CM4=shift(1, 127)-1); centerlift(Mod(2, prime(n))^CM4)
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CROSSREFS
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KEYWORD
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sign,fini
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AUTHOR
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STATUS
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approved
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