OFFSET
1,1
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
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 123 terms from Ray Chandler)
EXAMPLE
2523 = 3*29*29, 2525 = 5*5*101, 2527 = 7*19*19, 2529 = 3*3*281.
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
PROG
(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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Ray Chandler, Jun 24 2019
STATUS
approved