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The maximum length of a string of consecutive primes is 2: that is, the primes 2, 3. This is easily proven, as no even number other than 2 is prime.

In contrast, consider the set of numbers which are a product of exactly two primes (they don't need to be distinct). This set begins 4, 6, 9, 10, 12 ... It goes on to include the numbers 33, 34, 35 - a run of three consecutive integers. Is this the longest consecutive run of such numbers?

My conjecture: that each set of numbers which are the product of exactly n primes contains one and only one run of consecutive numbers which is n+1 long.

No idea whether this is true! Thoughts, anyone?

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    $\begingroup$ Out of four consecutive numbers, one is divisible by $4$. It follows that we cannot have four consecutive numbers that are products of $2$ primes. $\endgroup$ Commented Jun 11, 2016 at 1:46
  • $\begingroup$ They don't need to be distinct right? Like 25 counts as a semiprime $\endgroup$
    – JasonM
    Commented Jun 11, 2016 at 1:52
  • $\begingroup$ Have you tried to find a run of four consecutive numbers that are the product of three primes? Are you requiring the primes to be distinct? $\endgroup$ Commented Jun 11, 2016 at 2:03
  • $\begingroup$ Good clarification. The primes don't need to be distinct. $\endgroup$ Commented Jun 11, 2016 at 3:26
  • $\begingroup$ you should consider $\omega(n) = \sum_{p | n} 1$ instead, the number of distinct primes factors of $n$, without caring of the multiplicity $\endgroup$
    – reuns
    Commented Jun 11, 2016 at 5:56

1 Answer 1

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The limits each time are defined by the powers of $2$. So a consecutive set of numbers which each have exactly two prime factors (not necessarily distinct) will be bounded by the need to avoid multiples of $2^4=4$ (given that including $4$ itself is not an option).

And a consecutive set of numbers which have exactly three prime factors (not necessarily distinct) will be bounded by the need to avoid multiples of $2^3=8$. So we can possibly expect a run of $7$ such numbers, and the smallest such run starts at $211763$: $$\begin{array}{c|c} n &f_1 & f_2 & f_3 \\ \hline 211673 & 7 & 11 & 2749 \\ 211674 & 2 & 3 & 35279 \\ 211675 & 5 & 5 & 8467 \\ 211676 & 2 & 2 & 52919 \\ 211677 & 3 & 37 & 1907 \\ 211678 & 2 & 109 & 971 \\ 211679 & 13 & 19 & 857 \\ \end{array}$$

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  • $\begingroup$ A decent long run of numbers with four prime factors would require inclusion of a number $8$ times a prime. The best run under $10$ million is a set of nine consecutive numbers starting at $3405122$. $\endgroup$
    – Joffan
    Commented Jun 13, 2016 at 19:27
  • $\begingroup$ Great job finding such a run - how did you do it? Are there more such runs of 7? Will a run of 15 also exists for 4-prime numbers? $\endgroup$ Commented Jun 14, 2016 at 0:17
  • $\begingroup$ @ThomasDelaney pretty much a straight search for the 7-run of 3-factor numbers. There are 61 such sequences under 10 million. I 'm guessing there will be a 15-run of 4-factor numbers, but also that it will require a very large number space to allow such a coincidence to occur. The 9-run at 3405122 is the only one below 10 million. $\endgroup$
    – Joffan
    Commented Jun 14, 2016 at 0:31

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