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A308736
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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.
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3
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2523, 3112819, 5656019, 10132171, 12167825, 16639567, 25302173, 31995475, 35158921, 37334419, 43890719, 44816821, 47715269, 53548223, 55534523, 90526075, 90533525, 127558319, 142929025, 143167073, 144989575, 147182225
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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2523 = 3*29*29, 2525 = 5*5*101, 2527 = 7*19*19, 2529 = 3*3*281.
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MATHEMATICA
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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
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PROG
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(Python)
from sympy import factorint
A308736_list, n, mlist = [], 3, [False]*4
if mlist[0] and mlist[1] and mlist[2] and mlist[3]:
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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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