Sequence of 4 Consecutive Square-Free Triplets

Theorem

The following sets of $4$ consecutive triplets of integers, with one integer between each triplet, are square-free:

$29, 30, 31; 33, 34, 35; 37, 38, 39; 41, 42, 43$

$101, 102, 103; 105, 106, 107; 109, 110, 111; 113, 114, 115$

Proof

Note that $32, 36, 40$ and $104, 108, 112$ are all divisible by $4 = 2^2$, so are by definition not square-free.

Then inspecting each number in turn:

\(\ds 29\)

\(\)

\(\ds \)

is prime

\(\ds 30\)

\(=\)

\(\ds 2 \times 3 \times 5\)

and so is square-free

\(\ds 31\)

\(\)

\(\ds \)

is prime

\(\ds 33\)

\(=\)

\(\ds 3 \times 11\)

and so is square-free

\(\ds 34\)

\(=\)

\(\ds 2 \times 17\)

and so is square-free

\(\ds 35\)

\(=\)

\(\ds 5 \times 7\)

and so is square-free

\(\ds 37\)

\(\)

\(\ds \)

is prime

\(\ds 38\)

\(=\)

\(\ds 2 \times 19\)

and so is square-free

\(\ds 39\)

\(=\)

\(\ds 3 \times 13\)

and so is square-free

\(\ds 41\)

\(\)

\(\ds \)

is prime

\(\ds 42\)

\(=\)

\(\ds 2 \times 3 \times 7\)

and so is square-free

\(\ds 43\)

\(\)

\(\ds \)

is prime

\(\ds 101\)

\(\)

\(\ds \)

is prime

\(\ds 102\)

\(=\)

\(\ds 2 \times 3 \times 17\)

and so is square-free

\(\ds 103\)

\(\)

\(\ds \)

is prime

\(\ds 105\)

\(=\)

\(\ds 3 \times 5 \times 7\)

and so is square-free

\(\ds 106\)

\(=\)

\(\ds 2 \times 53\)

and so is square-free

\(\ds 107\)

\(\)

\(\ds \)

is prime

\(\ds 109\)

\(\)

\(\ds \)

is prime

\(\ds 110\)

\(=\)

\(\ds 2 \times 5 \times 11\)

and so is square-free

\(\ds 111\)

\(=\)

\(\ds 3 \times 37\)

and so is square-free

\(\ds 113\)

\(\)

\(\ds \)

is prime

\(\ds 114\)

\(=\)

\(\ds 2 \times 3 \times 19\)

and so is square-free

\(\ds 115\)

\(=\)

\(\ds 5 \times 27\)

and so is square-free

$\blacksquare$

Also see

• Sequence of Square-Free Triplets

This sequence is A007675 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $29$