OFFSET
1,1
COMMENTS
The terms are not congruent to 1 (mod 10).
The sequence contains no Mersenne prime p=2^t-1. Since p > 79, t is an odd prime and p+2 = 2^t+1 is divisible by 3. So, since 2^t+1 should be square, 2^t+1 is divisible by 9, i.e., (2^t+1)/3 == 0 (mod 3). (1)
Note that either t=6k+1 or t=6m+5. In each case, (1) is impossible.
Indeed, if t=6k+1, then (2^t+1)/3 = (2*(4^k)^3+1)/3 = (2*(3+1)^(3*k)+1)/3 == (2*binomial(3*k,1)*3+2+1)/3 == 1(mod 3), and analogously in case t=6*m+5, (2^t+1)/3 == 2 (mod 3): a contradiction.
LINKS
Peter J. C. Moses and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Moses)
FORMULA
Trivially a(n) >> n^2 log^2 n/(log log n)^2. - Charles R Greathouse IV, May 13 2015
EXAMPLE
Prime 79 is in the sequence because sqrt(79+2) = 9 = 3*3 which is semiprime.
Prime 1223 is in the sequence because sqrt(1223+2) = 35 = 5*7 which is semiprime.
MATHEMATICA
Select[Prime@Range@18000, PrimeOmega[Sqrt[#+2]]==2&]//Quiet (* Ivan N. Ianakiev, May 13 2015 *)
PROG
(PARI) issemi(n)=bigomega(n)==2
is(n)=isprime(n) && issquare(n+2, &n) && issemi(n) \\ Charles R Greathouse IV, May 13 2015
(PARI) list(lim)=my(v=List(), k=sqrt(lim+2), t); forprime(p=2, sqrt(k), forprime(q=p, k\p, if(isprime(t=(p*q)^2-2), listput(v, t)))); Set(v) \\ Charles R Greathouse IV, May 13 2015
(Perl) use ntheory ":all"; forprimes { say if is_power($_+2, 2) && scalar(factor(sqrtint($_+2)))==2 } 1e7; # Dana Jacobsen, May 13 2015
(Perl) use ntheory ":all"; sub list { my($lim, $k, $t, $p, %v)=shift; $k=sqrt($lim+2); forprimes { $p=$_; forprimes { $t=($p*$_)**2-2; $v{$t}++ if is_prime($t); } $p, int($k/$p); } int(sqrt($k)); my @v=sort{$a<=>$b} keys %v; @v; } say for list(1e10); # Translation of PARI, Dana Jacobsen, May 13 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, May 13 2015
EXTENSIONS
More terms from Peter J. C. Moses, May 13 2015
STATUS
approved