

A284278


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.


2



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

1,1


COMMENTS

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*p1 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*P1 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


LINKS



EXAMPLE

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.


MATHEMATICA

a[1]:=2;
a[n_?EvenQ]:=n/2+2;
a[n_?OddQ]:=If[PrimeQ[(n+1)/2+1], n+2, 2];


PROG

(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


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



