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%I #28 Feb 24 2016 11:29:21
%S 2,5,11,17,23,41,47,83,89,101,137,167,179,251,257,353,359,401,461,503,
%T 641,719,809,821,881,941,1013,1097,1151,1283,1361,1409,1433,1439,1601,
%U 1619,1871,2027,2069,2351,2531,2657,2663,2741,2789,2819,2879,3203,3209,3581
%N The elements of the set P3 in ascending order.
%C P3 is the largest set of primes satisfying the conditions: (1) 3 is not in P3; (2) a prime p>3 is in P3 iff all prime divisors of p-1 are in P3.
%C P3 is also the set of all primes p for which the Pratt tree for p has no node labeled 3.
%C It is conjectured that this sequence is infinite.
%H Alois P. Heinz, <a href="/A220813/b220813.txt">Table of n, a(n) for n = 1..10000</a>
%H K. Ford, S. Konyagin and F. Luca, <a href="http://arxiv.org/abs/0904.0473">Prime chains and Pratt trees</a>, Geom. Funct. Anal., 20 (2010), pp. 1231-1258 (arXiv:0904.0473 [math.NT]).
%H Kevin Ford, <a href="http://arxiv.org/abs/1212.3498">Sieving by very thin sets of primes, and Pratt trees with missing primes</a>, arXiv preprint arXiv:1212.3498 [math.NT], 2012-2013.
%F Ford proves that a(n) >> n^k for some k > 1. "It appears" that k can be taken as 1.612. - _Charles R Greathouse IV_, Dec 26 2012
%e 11 is in P3, because 11-1 = 2*5 and 2, 5 are in P3.
%p with(numtheory):
%p P3:= proc(n) P3(n):= `if`(n<1, {}, P3(n-1) union {a(n)}) end:
%p a:= proc(n) option remember; local p;
%p if n<3 then [2, 5][n]
%p else p:=a(n-1);
%p do p:= nextprime(p);
%p if factorset(p-1) minus P3(n-1) = {} then break fi
%p od; p
%p fi
%p end:
%p seq(a(n), n=1..70); # _Alois P. Heinz_, Dec 26 2012
%t P3 = {2, 5}; For[p=11, p<4000, p=NextPrime[p], If[ AllTrue[ FactorInteger[ p-1][[All, 1]], MemberQ[P3, #]&], AppendTo[P3, p]]]; P3 (* _Jean-François Alcover_, Feb 24 2016 *)
%o (PARI) P(k,n)=if(n<=k, n<k, my(f=factor(n-1)[,1]); for(i=1, #f, if(!P(k, f[i]), return(0))); 1)
%o is(n)=isprime(n) && P(3,n) \\ _Charles R Greathouse IV_, Dec 26 2012
%Y Cf. A220814, A220815.
%K nonn
%O 1,1
%A _Franz Vrabec_, Dec 22 2012
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