How can we generalize any theorem for finding \(n\) consecutive positive composite numbers?

How can we generalize any theorem for finding \(n\) consecutive positive composite numbers?

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TopNewestFor \(n \ge 1\) 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. – Brian Charlesworth · 1 year, 10 months ago

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– Abhijeet Verma · 1 year, 10 months ago

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

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. – Brian Charlesworth · 1 year, 10 months ago

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– Abhijeet Verma · 1 year, 10 months ago

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

@Brian Charlesworth Sir, please help again! – Abhijeet Verma · 1 year, 10 months ago

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