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A215888
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Numbers k such that 2^k - 1 - Sum_{prime p<k} 2^p is prime.
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1
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2, 3, 4, 5, 6, 7, 8, 21, 25, 33, 40, 58, 158, 226, 240, 295, 465, 730, 1774, 2184, 2352, 3279, 4221, 6340, 6373, 6374, 17822, 18591, 19456, 24410, 32085, 64534
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OFFSET
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1,1
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COMMENTS
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The primes of this form can be written with k binary digits, which are 0 in prime positions and 1 in all other positions. Here, "position p" means the digit with value 2^p (so the LSB is position 0). See A215891 for a variant where the rightmost bit is position 1.
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LINKS
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EXAMPLE
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k=2, 3 and 4 are in the sequence because 2^2 - 1 = 2^3 - 1 - 2^2 = 2^4 - 1 - 2^2 - 2^3 = 3 is prime. (For k=3 and k=4, since the uppermost digits are set to zero, one gets the same prime as for k=2 which is in fact only of binary length 2.)
k=5 and k=6 are in the sequence because 2^5 - 1 - 2^2 - 2^3 = 2^6 - 1 - 2^2 - 2^3 - 2^5 = 19 is prime. (For k=6, 2^5 cancels the uppermost bit so again we have only 5 binary digits.)
k=7 and k=8 are in the sequence because 2^7 - 1 - 2^2 - 2^3 - 2^5 = 2^8 - 1 - 2^2 - 2^3 - 2^5 - 2^7 = 83 is prime. (Here again, 2^7 cancels the uppermost bit so we have only 7 binary digits.) One can see that whenever a term a(n) of the sequence is the larger of a twin prime pair, then a(n+1) = a(n)+1 is also in the sequence and corresponds to the same a(n)-digit prime.
k=21 is in the sequence, because 2^21 - 1 - 2^2 - ... - 2^19 = 1431379 is prime (and has binary length 21).
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MATHEMATICA
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isA215888[n_] := PrimeQ[2^n-1 - Sum[2^Prime[k], {k, 1, PrimePi[n-1]}]]; Reap[For[n = 1, n < 10^5, n++, If[isA215888[n], Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Dec 26 2013, after PARI *)
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PROG
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(PARI) is_A215888(N)=ispseudoprime(2^N-1-sum(n=1, primepi(N-1), 2^prime(n)))
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CROSSREFS
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KEYWORD
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nonn,hard,nice,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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