OFFSET
1,1
COMMENTS
Except for the second term, a(n)+1 is divisible by 6.
[Proof: a(n)=p is a prime, with p-1=q*r and two primes q<=r by definition. Omitting the special case p=2, p is odd, p+1 is even, so p+1=q*r+2 = 2(1+q*r/2). To show that p+1 is divisible by 6 we show that it is divisible by 2 and by 3; divisibility by 2 has already been shown in the previous sentence. (1+q*r/2 must be integer, so q*r/2 must be integer, so the smaller prime q of the semiprime must be q=2, so p=2*r+1. This shows that p=a(n) are a subset of A005383.) First subcase of the definition is that p+2 is also prime. Then p is a smaller twin prime and by a comment in A003627, p+1 is divisible by 3. Second subcase of the definition is that p+2 = s^2 with s a prime. s can be 3*k+1 or 3*k+2 --p=7 is the exception-- which leads to s^2 = 9*k^2+6*k+1 or s^2=9*k^2+12*k+4, so p+1 = 9*k^2+6*k or 9*k^2+12*k+3, and in both cases p+1 is divisible by 3.]
In consequence, except for the first three terms, first differences a(n+1)-a(n) are also divisible by 6.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
semiPrimeQ[n_] := Plus @@ Last /@ FactorInteger@n == 2; fQ[n_] := Block[{fi = FactorInteger@n}, Length@ fi == 1 && fi[[1, 2]] == 1 || fi[[1, 2]] == 2]; Select[ Prime@ Range@ 1293, semiPrimeQ[ # - 1] && fQ[ # + 2] &] (* Robert G. Wilson v, Nov 06 2010 *)
Select[Prime[Range[2000]], PrimeOmega[#-1]==2&&Or@@PrimeQ[{#+2, Sqrt[ #+2]}]&] (* Harvey P. Dale, Aug 12 2012 *)
PROG
(Magma) [ p: p in PrimesInInterval(3, 15000) | &+[ k[2]: k in Factorization(p-1) ] eq 2 and (IsPrime(p+2) or (q^2 eq p+2 and IsPrime(q) where q is Isqrt(p+2))) ]; // Klaus Brockhaus, Nov 03 2010
CROSSREFS
KEYWORD
nonn
AUTHOR
Giovanni Teofilatto, Nov 01 2010
EXTENSIONS
Corrected (29 removed) and extended by Klaus Brockhaus, Robert G. Wilson v and R. J. Mathar, Nov 03 2010
STATUS
approved