A308736 Numbers n such that n, n+2, n+4, n+6 are of the form p^2*q where p and q are distinct primes.

2523, 3112819, 5656019, 10132171, 12167825, 16639567, 25302173, 31995475, 35158921, 37334419, 43890719, 44816821, 47715269, 53548223, 55534523, 90526075, 90533525, 127558319, 142929025, 143167073, 144989575, 147182225

Offset: 1

Author

Ray Chandler, Jun 24 2019

Keywords

• nonn

Comments

All terms are odd. Proof: if n is even then out of the 4 numbers n, n+2, n+4, n+6, 2 of them must be either both of the form 2*p^2, 2*q^2, or both of the form 4*p, 4*q. In either case, for p != q and p, q prime, the difference between these 2 numbers are more than 6, reaching a contradiction. - Chai Wah Wu, Jun 24 2019

Examples

			2523 = 3*29*29, 2525 = 5*5*101, 2527 = 7*19*19, 2529 = 3*3*281.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 123 terms from Ray Chandler)

Crossrefs

Cf. A074173, A308735.

Programs

Mathematica

psx = Table[{0}, {7}]; nmax = 150000000; n = 1; lst = {};
While[n < nmax, n++;
  psx = RotateRight[psx];
  psx[[1]] = Sort[Last /@ FactorInteger[n]];
  If[Union[{psx[[1]], psx[[3]], psx[[5]], psx[[7]]}] == {{1, 2}}, AppendTo[lst, n - 6]];];
lst

Python

from sympy import factorint
A308736_list, n, mlist = [], 3, [False]*4
while len(A308736_list) < 100:
    if mlist[0] and mlist[1] and mlist[2] and mlist[3]:
        A308736_list.append(n)
    n += 2
    f = factorint(n+6)
    mlist = mlist[1:] + [(len(f),sum(f.values())) == (2,3)] # Chai Wah Wu, Jun 24 2019, Jan 03 2022.