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A220813 The elements of the set P3 in ascending order. 3
2, 5, 11, 17, 23, 41, 47, 83, 89, 101, 137, 167, 179, 251, 257, 353, 359, 401, 461, 503, 641, 719, 809, 821, 881, 941, 1013, 1097, 1151, 1283, 1361, 1409, 1433, 1439, 1601, 1619, 1871, 2027, 2069, 2351, 2531, 2657, 2663, 2741, 2789, 2819, 2879, 3203, 3209, 3581 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
P3 is also the set of all primes p for which the Pratt tree for p has no node labeled 3.
It is conjectured that this sequence is infinite.
LINKS
K. Ford, S. Konyagin and F. Luca, Prime chains and Pratt trees, Geom. Funct. Anal., 20 (2010), pp. 1231-1258 (arXiv:0904.0473 [math.NT]).
Kevin Ford, Sieving by very thin sets of primes, and Pratt trees with missing primes, arXiv preprint arXiv:1212.3498 [math.NT], 2012-2013.
FORMULA
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
EXAMPLE
11 is in P3, because 11-1 = 2*5 and 2, 5 are in P3.
MAPLE
with(numtheory):
P3:= proc(n) P3(n):= `if`(n<1, {}, P3(n-1) union {a(n)}) end:
a:= proc(n) option remember; local p;
if n<3 then [2, 5][n]
else p:=a(n-1);
do p:= nextprime(p);
if factorset(p-1) minus P3(n-1) = {} then break fi
od; p
fi
end:
seq(a(n), n=1..70); # Alois P. Heinz, Dec 26 2012
MATHEMATICA
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 *)
PROG
(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)
is(n)=isprime(n) && P(3, n) \\ Charles R Greathouse IV, Dec 26 2012
CROSSREFS
Sequence in context: A113426 A078894 A086319 * A217303 A053033 A136244
KEYWORD
nonn
AUTHOR
Franz Vrabec, Dec 22 2012
STATUS
approved

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Last modified May 13 00:07 EDT 2024. Contains 372497 sequences. (Running on oeis4.)