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A232565
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a(n) is the smallest k such that 2^(2^n) - 2^k - 1 is prime, or -1 if no such k exists.
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3
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0, 1, 2, 4, 2, 8, 18, 76, 32, 151, 692, 592, 154, 580, 27365, 11267
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OFFSET
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1,3
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COMMENTS
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Crocker showed that 2^(2^n) - 1 - 2^a - 2^b is not prime (with n > 2) if a and b are distinct. This sequence demonstrates that the theorem is sharp in the sense that distinctness is required.
If n > 2, then the (largest) prime P(n) = 2^(2^n)-2^a(n)-1 is a de Polignac number (A065381); i.e., P(n)-2^m is not prime. It seems that if n > 6, then |P(n)-2^m| is composite for every natural m and P(n)*2^m-1 is composite (by the dual Riesel conjecture). So if n > 6, then the prime P(n) may be a Riesel number (A182296). For example, the prime P(7) = 2^(2^7)-(2^18+1) is the first candidate (note that 2^18+1 is the smallest de Polignac number of form 2^k+1). Also, by Crocker's theorem, the smallest number of form 2^(2^n)-2^m-1, namely 2^(2^n-1)-1 is a de Polignac number (A006285) and for n > 6 may be a dual Riesel number (A101036). For example, the double Mersenne prime 2^(2^7-1)-1 probably is a dual Riesel number. It is not known whether these are Riesel numbers with a covering set. - Thomas Ordowski, Jan 24 2024
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LINKS
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PROG
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(PARI) a(n)=my(N=2^2^n-1); for(a=1, 2^n-1, if(ispseudoprime(N-2^a), return(a))); 0
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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