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A283191
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Prime numbers p > 2 such that (2^p - 5)/3 is prime.
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0
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OFFSET
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1,1
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COMMENTS
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Let W = (2^p - 5)/3 and s = (W+1)/(2*p), then 5^s == 2 (mod W) for terms 1..9.
Subsequence of 7, 13, 19, 31, 51, 55, 85, 111, 319, 373,.. which are numbers m such that (2^m-5)/3 is prime. - R. J. Mathar, Mar 05 2017
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LINKS
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MATHEMATICA
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Select[Prime@ Range[2, 1000], PrimeQ[(2^# - 5)/3] &] (* Michael De Vlieger, Mar 03 2017 *)
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PROG
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(PARI)
forprime(p=3, 30000, W= (2^p-5)/3; if(ispseudoprime(W), print1(p, ", ")))
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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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STATUS
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approved
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