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A108294
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a(n) = the least prime p such that p-6n-3 is a power of 2 and 2p-6n-3 is prime, or -1 if no such prime exists.
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2
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5, 11, 17, 29, 29, 37, 41, 53, 59, 59, 67, 71, 79, 89
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OFFSET
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0,1
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COMMENTS
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The value of a(14) is presently unknown.
If a(n) is not -1, it is 2^k+6n+3 where both 2^k+6n+3 and 2^(k+1)+6n+3 are prime, and k is the least number with this property.
Heuristically, the probability that 2^k+6n+3 is prime should be on the order of k^(-1), and the probability that both 2^k+6n+3 and 2^(k+1)+6n+3 are prime should be on the order of k^(-2). Thus we should expect only finitely many such k for a given n, and if it does not occur for fairly small k it is unlikely to ever occur. However, this is not a proof.
If it is not -1, a(14) > 2^10000+87. (End)
Values a(14) through a(60) for 2 <= p <= prime(10^6): {-1, 97, 101, 107, 239, 149, 127, 193, 137, 149, 149, 281, 163, 173, 32939, 179, 191, 191, 197, -1, 223, -1, 223, 227, 359, 239, -1, 281, 263, 269, 269, 277, 281, 317, 419, 809, 307, 311, 331, 337, -1, 397, -1, 347, 359, 373, 491}. Unknown values (-1) at n = 14, 33, 35, 40, 54, 56. - Michael De Vlieger, Aug 04 2016
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LINKS
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MAPLE
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# values of -1 from this function indicate either a(n)=-1 or k > 10000
f:= proc(n) local j, k, p, q, state, goodk;
state:= 0;
goodk:= select(t -> igcd(6*n+3+2^t, 6*n+3+2^(t+1), 5*7*11*13*17)=1, [$0..119]);
for k from 1 to 10000 do
if (state = 0 and not member(k mod 120, goodk)) then state:= 0; next fi;
p:= 6*n+3+2^k;
if isprime(p) then
state:= state+1;
if state = 2 then return p - 2^(k-1) fi;
else
state:= 0;
fi;
od;
-1;
end proc:
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MATHEMATICA
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Table[SelectFirst[Prime@ Range[10^5], IntegerQ@ Log2[# - 6 n - 3] && PrimeQ[2 # - 6 n - 3] &], {n, 0, 60}] /. k_ /; MissingQ@ k -> -1 (* Michael De Vlieger, Aug 04 2016, Version 10.2 *)
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CROSSREFS
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KEYWORD
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nonn,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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