213

Since August 1 is the 213th day of the year (2023 not being a leap year), this seems like a good time to share an interesting fact about the number 213.

It involves semiprimes, so let’s review what a semiprime is: a product of two distinct prime numbers. 213 is a semiprime because it is the product of 3 × 71 and each of those numbers is prime.

The first few semiprimes are 6, 10, 14, and 15 (2×3, 2×5, 2×7, and 3×5 respectively.) There are infinitely many pairs of semiprimes, such as 14 and 15, 21 and 22, and 33 and 34. And there are infinitely many sets of three consecutive semiprimes, such as 33/34/35, 85/86/87, and 93/94/95.

A set of four consecutive semiprimes is not possible, however. Every fourth number is divisible by 4, and if a number is divisible by 4 it cannot be a semiprime, since a semiprime contains only distinct primes as factors, and 4 is 2 × 2, and the 2s are not distinct.

However, if we skip numbers divisible by 4, we can find longer runs of semiprimes. 213 is the smallest integer that begins a run of 6 out of 7 semiprimes, and it is the only number less than 100,000 to start such a run. Here is the prime factorization of 213 and the next few semiprimes that follow it:

Number	Prime factorization	Semiprime?
213	3 × 71	✓
214	2 × 107	✓
215	5 × 43	✓
216	23 × 33	—
217	7 × 31	✓
218	2 × 109	✓
219	3 × 73	✓
220	22 × 5 × 11	—
221	13 × 17	✓
222	2 × 3 × 37	—

In addition to starting a run of 6 semiprimes out of 7, 213 also begins a run of 7 semiprimes out of 9, ending with 221. A run of 8 semiprimes out of 10 consecutive integers is not possible since the tenth number in the sequence will be divisible by both 2 and 3.

Beyond 213

However, it is possible to get a run of 8 semiprimes out of 11, although not with the 213 sequence. Instead, we have to look at some of the other sequences that contain 6 semiprimes out of 7 consecutive integers. We can look to the Online Encyclopedia of Integer Sequences, sequence A242804 specifically:

A242804: Integers k such that each of k, k+1, k+2, k+4, k+5, k+6 is the product of two distinct primes:

213, 143097, 194757, 206133, 273417, 684897, 807657, 1373937, 1391757, 1516533, 1591593, 1610997, 1774797, 1882977, 1891761, 2046453, 2051493, 2163417, 2163957, 2338053, 2359977, 2522517, 2913837, 3108201, 4221753

Let’s take a look at 143097, the second entry in the list. In addition to the six semiprimes mentioned in the description (k, k+1, etc.), factoring k-2 (143095) and k+8 (143105) shows that those two are semiprimes as well:

Number	Prime factorization	Semiprime?
143095	5 × 28619	✓
143096	23 × 31 × 577	—
143097	3 × 47699	✓
143098	2 × 71549	✓
143099	11 × 13009	✓
143100	22 × 33 × 52 × 53	—
143101	7 × 20443	✓
143102	2 × 71551	✓
143103	3 × 47701	✓
143104	28 × 13 × 43	—
143105	5 × 28621	✓

So that’s 8 semiprimes out of 11 consecutive integers. What other numbers from the A242804 sequence have this property?

It turns out 143097 is the only number under a million that does. However, there are larger numbers from that list that are part of such a sequence. But rather than list those numbers, let’s list the numbers that actually start an 8-of-11 run, which are two less than the numbers shown in A242804. Here are the first twenty:

143095, 2046451, 5108791, 5247139, 5606239, 10389199, 10496839, 10630831, 12701299, 16917511, 17193451, 30398539, 33949651, 40516555, 40912951, 48634951, 57758899, 59023939, 64430851, 75184735

Like sequence A242804, the second element of this sequence is an order of magnitude larger than the first.

Going further

Can we do even better? Can we find some more semiprimes adjacent to the entries in A364751? There are at least two which have a neighboring semiprime in both the k-2 and k+12 positions:

Position	Number	Prime factorization	Number	Prime factorization
k-2	10630829	11 × 966439	64430849	7 × 9204407
k	10630831	17 × 625343	64430851	2179 × 29569
k+2	10630833	3 × 3543611	64430853	3 × 21476951
k+3	10630834	2 × 5315417	64430854	2 × 32215427
k+4	10630835	5 × 2126167	64430855	5 × 12886171
k+6	10630837	7 × 1518691	64430857	653 × 98669
k+7	10630838	2 × 5315419	64430858	2 × 32215429
k+8	10630839	3 × 3543613	64430859	3 × 21476953
k+10	10630841	13 × 817757	64430861	11 × 5857351
k+12	10630843	3163 × 3361	64430863	7 × 9204409

This gives us 10 semiprimes out of 15 consecutive integers. The table skips k-1 and k+11 because they are divisible by 6, and does not show k-3, k+1, k+5, k+9, or k+13 because they are divisible by 4. Extending it to include k-4 or k+14 would not help us find more neighboring semiprimes since both k-4 and k+14 would be divisible by 9.

It appears that the runs containing 10,630,831 and 64,430,851 are the only such 10-out-of-15 runs that contain numbers less than 100,000,000. 10 semiprimes is the highest number we can squeeze out of 15 consecutive integers, and with only 2 such sequences out of the first 100 million integers, such runs are a quite rare sight.