# Consecutive $$n$$ composite numbers

How can we generalize any theorem for finding $$n$$ consecutive positive composite numbers?

Note by Abhijeet Verma
2 years, 9 months ago

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For $$n \ge 1$$ consider the $$n$$-term sequence $$(n + 1)! + 2, (n + 1)! + 3, (n + 1)! + 4, ..... , (n + 1)! + (n + 1).$$

Then the $$k$$th term is divisible by $$k + 1$$ for $$1 \le k \le n,$$ and thus all $$n$$ terms are composite.

- 2 years, 9 months ago

Are these terms the lowest possible numbers as well? I guess no, so can we generalize them for lowest possible terms?

- 2 years, 9 months ago

I'm not sure if there will be a formula for the lowest possible terms for a given $$n.$$ As you point out $$8,9,10$$ are the lowest for $$n = 3.$$ For $$n = 2$$ it would be $$8,9,$$ (which matches the sequence I mentioned above). For $$n = 4$$ it would be $$24,25,26,27$$ and for $$n = 5$$ it would be $$24,25,26,27,28.$$ For $$n = 6$$ it would be $$90,91,92,93,94,95$$ and for $$n = 7$$ it would be $$90,91,92,93,94,95,96.$$

Since prime gaps for primes in excess os $$2$$ are multiples of $$2$$ we will find that the lowest sequences for $$n = 2k$$ and $$n = 2k + 1$$ will have the same lowest term. Beyond that observation, I think it would be difficult to establish any general formula since the number of consecutive composites depends on the sizes of the prime gaps, (i.e., difference between successive primes), which don't follow any identifiable pattern.

- 2 years, 8 months ago

As if I have to find three consecutive positive composite numbers, the lowest are 8,9 and 10.

- 2 years, 9 months ago