A368785 Least of three consecutive primes p, q, r such that p + q, p + r, q + r and p + q + r all have the same number of prime divisors, counted with multiplicity.

1559, 4073, 5237, 5987, 12119, 14633, 24697, 29881, 29947, 30113, 32003, 41903, 45863, 60169, 64817, 67601, 69151, 71263, 73783, 77713, 78929, 79633, 86629, 88547, 91493, 95483, 96181, 108037, 109859, 110459, 111667, 125471, 132833, 133283, 140419, 142049, 160001, 165133, 170579, 171803, 171827, 171947

Offset: 1

Author

Zak Seidov and Robert Israel, Jan 05 2024

Keywords

• nonn

Comments

The number of prime divisors is at least 3, because p + q is even and not twice a prime.

Examples

			a(2) = 4073 is a term because 4073, 4079, 4091 are consecutive primes with
4073 + 4079 = 8152 = 2^3 * 1019,
4073 + 4091 = 8164 = 2^2 * 13 * 157,
4079 + 4091 = 8170 = 2 * 5 * 19 * 43, and
4073 + 4079 + 4091 = 12243 = 3 * 7 * 11 * 53
all have 4 prime divisors, counted with multiplicity.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001222, A368786.

Programs

Maple

R:= NULL: count:= 0:
p:= 2: q:= 3: r:= 5: v:= numtheory:-bigomega(q+r);
while count < 100 do
  p:= q; q:= r; r:= nextprime(r);
  w:= numtheory:-bigomega(q+r);
  if w = v and numtheory:-bigomega(p+r) = v and numtheory:-bigomega(p+q+r) = v then
    R:= R,p; count:= count+1;
    fi;
  v:= w;
od:
R;

Mathematica

Select[Partition[Prime[Range[16000]],3,1],Length[Union[PrimeOmega[Total/@Subsets[#,{2,3}]]]]==1&][[;;,1]] (* Harvey P. Dale, May 25 2025 *)