Triples of Consecutive Sphenic Numbers
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Theorem
The sequence of triplets of consecutive sphenic numbers starts:
- $\tuple {1309, 1310, 1311}, \tuple {1885, 1886, 1887}, \tuple {2013, 2014, 2015}, \ldots$
The sequence of the first elements is A066509 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
The sequence of the middle elements is A248202 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
Note that there cannot be quadruplets of such numbers, since one of the quadruplets must be divisible by $4$, making it non-sphenic.
We have:
\(\ds 1309\) | \(=\) | \(\ds 7 \times 11 \times 17\) | ||||||||||||
\(\ds 1310\) | \(=\) | \(\ds 2 \times 5 \times 131\) | ||||||||||||
\(\ds 1311\) | \(=\) | \(\ds 3 \times 19 \times 23\) | ||||||||||||
\(\ds 1885\) | \(=\) | \(\ds 5 \times 13 \times 29\) | ||||||||||||
\(\ds 1886\) | \(=\) | \(\ds 2 \times 23 \times 41\) | ||||||||||||
\(\ds 1887\) | \(=\) | \(\ds 3 \times 17 \times 37\) | ||||||||||||
\(\ds 2013\) | \(=\) | \(\ds 3 \times 11 \times 61\) | ||||||||||||
\(\ds 2014\) | \(=\) | \(\ds 2 \times 19 \times 53\) | ||||||||||||
\(\ds 2015\) | \(=\) | \(\ds 5 \times 13 \times 31\) |
hence each number above is sphenic.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1309$