

A079149


Primes p such that either p1 or p+1 has at most 2 prime factors, counted with multiplicity; i.e., primes p such that either bigomega(p1) <= 2 or bigomega(p+1) <= 2, where bigomega(n) = A001222(n).


6



2, 3, 5, 7, 11, 13, 23, 37, 47, 59, 61, 73, 83, 107, 157, 167, 179, 193, 227, 263, 277, 313, 347, 359, 383, 397, 421, 457, 467, 479, 503, 541, 563, 587, 613, 661, 673, 719, 733, 757, 839, 863, 877, 887, 983, 997, 1019, 1093, 1153, 1187, 1201, 1213, 1237
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OFFSET

1,1


COMMENTS

There are only 2 primes such that both p1 and p+1 have at most 2 prime factors  3 and 5. Proof: If p > 5 then whichever of p1 and p+1 is divisible by 4 has at least 3 prime factors.
Primes which not are the sum of two consecutive composite numbers.  JuriStepan Gerasimov, Nov 15 2009
Taking consecutive composites a and b with a > b, form a + b and a^2  b^2. Considering the subset where a + b = a^2  b^2, this sequence is the primes that are NOT encountered. For example, with 8 and 9, a + b = a^2  b^2 = 17. The next prime is encountered from 14 and 15, a + b = a^2  b^2 = 29. Thus, prime 23 has been missed, and it is in this sequence.  Bill McEachen, Dec 25 2013


LINKS

Table of n, a(n) for n=1..53.


MATHEMATICA

Select[Prime[Range[500]], MemberQ[PrimeOmega[{#1, #+1}], 2]&] (* Harvey P. Dale, Sep 04 2011 *)


PROG

(PARI) s(n) = {sr=0; ct=0; forprime(x=2, n, if(bigomega(x1) < 3  bigomega(x+1) < 3, print1(x" "); sr+=1.0/x; ct+=1; ); ); print(); print(ct" "sr); } \\ Lists primes p<=n such that p+1 has at most 2 prime factors.


CROSSREFS

Union of A079147 and A079148. Cf. A060254, A079152.
Sequence in context: A291691 A178576 A038970 * A024694 A024320 A265885
Adjacent sequences: A079146 A079147 A079148 * A079150 A079151 A079152


KEYWORD

easy,nonn


AUTHOR

Cino Hilliard, Dec 27 2002


STATUS

approved



