A323498 Primes p such that p - k and p + k have the same number of prime factors (with multiplicity), for k = 1..6.

2131991, 2917927, 3776273, 4742407, 6853409, 16850609, 21789233, 24095791, 24810251, 26316233, 27470537, 27667529, 28962127, 29896439, 30949327, 31289527, 36123853, 36443893, 38824913, 40941233, 41660009, 42533551, 44233193, 45868967, 48313567, 49265009, 51135991

Offset: 1

Author

Zak Seidov, Jan 16 2019

Keywords

• nonn

Comments

At least one of p - k and p + k must be composite for each k in for k = 1..5.

Proof: If k = 3 then p - k and p + k are even. If k isn't three then exactly one of p - k, p and p + k is divisible by 3. QED. - David A. Corneth, Jan 18 2019

Examples

			For p = 2131991 is in the sequence because for k=1, p - 1 = 2*5*7*7*9*229 and p + 1 = 2*2*2*3*3*29611 are both 6-almost primes, for k=2, p - 2 = 3*710663 and p + 2 = 29*73517 are both semiprimes, etc.
		

Crossrefs

Cf. A115103 (k=1), A323536 (k=7), A323537 (k=8).

Programs

PARI

upto(n) = {my(res = List(), q = 5); forprime(p = 7, n, t = 1; for(m = 1, 2, for(i = 0, 2, if(bigomega(p + 2*i + m) != bigomega(p - 2*i - m), t = 0; next(2) ) ) ); if(t == 1, listput(res, p)); q = p; ); res } \\ David A. Corneth, Jan 17 2019

PARI

is(n) = if(!isprime(n) || n < 7, return(0)); for(k = 1, 6, if(bigomega(n + k) != bigomega(n - k), return(0))); 1 \\ David A. Corneth, Jan 17 2019

Perl

use ntheory ':all'; for (my($p,$k)=(2,6); $p <= 10**7; $p = next_prime($p)) { print "$p\n" if vecall {factor($p-$) == factor($p+$)} 1..$k } # Daniel Suteu, Jan 17 2019

Extensions

a(23)-a(27) from David A. Corneth, Jan 17 2019