A189759 Numbers pqr such that pq + pr + qr is prime, where p, q, and r are primes.

30, 42, 66, 70, 78, 105, 114, 130, 154, 165, 174, 182, 222, 231, 238, 246, 255, 273, 282, 285, 286, 310, 318, 345, 357, 366, 370, 385, 399, 418, 430, 434, 442, 455, 465, 474, 483, 494, 498, 518, 555, 561, 574, 582, 595, 602, 609, 618, 642, 645, 651, 663, 665

Offset: 1

Author

T. D. Noe, Apr 27 2011

Keywords

• nonn

Comments

The number pq+pr+qr is prime only if p, q, and r are distinct. The primes of form pq+pr+qr are in A087054. A prime may have multiple representations as pq+pr+qr; for example, 2*3*13 and 3*5*7 both produce the prime 71.

As mentioned by Ufnarovski and Ahlander, if pq+pr+qr is prime, then the arithmetic derivative (A003415) of pqr is that prime. They conjecture that this sequence and A087054 are infinite.

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.

Crossrefs

Cf. A003415, A087054, A356475.

Programs

Mathematica

pqr[nn_] := Module[{p=Prime[Range[PrimePi[nn/6]+1]],i,j,k,n,prod}, Sort[Reap[i=0; While[i++; p[[i]]p[[i+1]]p[[i+2]] <= nn, j=i; While[j++; p[[i]]p[[j]]p[[j+1]] <= nn, k=j; While[k++; prod=p[[i]]p[[j]]p[[k]]; prod <= nn, n=p[[i]]p[[j]]+p[[i]]p[[k]]+p[[j]]p[[k]]; If[PrimeQ[n], Sow[prod]]]]]][[2,1]]]]; pqr[1000]
Take[Union[Times@@@Select[Subsets[Prime[Range[30]],{3}],PrimeQ[ Total[ Times@@@Subsets[#,{2}]]]&]],60](* Harvey P. Dale, Dec 29 2011 *)