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A284278
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a(1)=2; for n >= 1, if n+2 is prime then a(2*n+1) = 3*n + 2 and a(2*n) = n + 2, otherwise all terms are 2.
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2
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2, 3, 5, 4, 2, 5, 9, 6, 2, 7, 13, 8, 2, 9, 2, 10, 2, 11, 21, 12, 2, 13, 25, 14, 2, 15, 2, 16, 2, 17, 33, 18, 2, 19, 37, 20, 2, 21, 2, 22, 2, 23, 45, 24, 2, 25, 2, 26, 2, 27, 2, 28, 2, 29, 57, 30, 2, 31, 61, 32, 2, 33, 2, 34, 2, 35, 2, 36, 2, 37, 73, 38, 2, 39
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OFFSET
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1,1
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COMMENTS
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The sequence is motivated by A284172, by the message from B. Jubin dated Mar 23 2017 and by the classic open problem of showing that there are infinitely many primes p for which 2*p-1 is also prime. If there were only finitely many such primes, then there would be a place where this sequence is generated by the same rule as A284172.
The sequence of the first differences begins 1, 2, -1, -2, 3, 4, -3, -4, 5, 6, -5, -6, 7, -7, 8, -8, 9, 10, -9, -10, 11, 12, -11, -12, 13, -13, 14, -14, 15, 16, -15, -16, 17, 18, ...
From the definition it easily follows that, for a positive x, the sequence contains roughly equal numbers of prime and composite terms <= x.
A conditional property: if there is a maximal prime P such that 2*P-1 is also prime, then for n > P, every pair (a(2*n), a(2*n+1)) contains one prime and one composite. Indeed, if n+2 is prime, then a(2*n) = n + 2 is prime, while a(2*n+1) = 2*n + 3 = 2*(n+2) - 1 is composite; if n+2 is composite, then a(2*n) = n + 2 is composite, while a(2*n+1) = 2 is prime. - Vladimir Shevelev, Mar 26 2017
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LINKS
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EXAMPLE
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For n=19, a(38) = a(2*19) = 19+2 = 21, a(39) = a(2*19+1) = 2, the latter since 19+2 is not prime;
for n=21, a(42) = a(2*21) = 21+2 = 23, a(43) = a(2*21+1) = 2*21+3 = 45 since 21+2 is prime.
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MATHEMATICA
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a[1]:=2;
a[n_?EvenQ]:=n/2+2;
a[n_?OddQ]:=If[PrimeQ[(n+1)/2+1], n+2, 2];
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PROG
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(PARI) a(n) = if(n<2, 2, if(n%2, if(isprime((n + 1)/2 + 1), n + 2, 2), (n/2 + 2))); \\ Indranil Ghosh, Mar 25 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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