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Is there any particular arithmetic progression of natural numbers in which the number of primes is greater than the number of composites till any term of the progression? (Either proven or conjectured). If I list the terms- a, a+d, a+2d, ......, m (where m = a + nd for some n). Then no matter which term I stop at, the number of primes encountered till then should be greater than the number of composites

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  • $\begingroup$ Sure: the sequence of all prime numbers (by increasing order, for instance). $\endgroup$ – José Carlos Santos Jul 4 '18 at 17:56
  • $\begingroup$ I should have framed it properly. Is there an A.P. like that? Actually I was working with the sequence 5, 11, 17,....., 6n-1. At 1139 the number of composites become greater $\endgroup$ – Pranali Jul 4 '18 at 18:00
  • $\begingroup$ Else, is there any conjecture about the existence of such a progression? $\endgroup$ – Pranali Jul 4 '18 at 18:08
  • $\begingroup$ It sure contains infinitely many prime numbers. But if we consider the first m terms of the progression (for any m) then the number of primes may not always be greater than number of composites $\endgroup$ – Pranali Jul 4 '18 at 18:12
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Mason Jul 4 '18 at 18:12
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No. It is known that $\pi(n)$, the number of primes not exceeding $n$, is asymptotically $\approx n/\ln n$. If an arithmetic progression with step $d$ would have a majority of primes, then we would have $$\liminf_{n\to\infty}\pi(n)/n\ge 1/(2d).$$

But $\pi/n\approx1/\ln n\to0$ as $n\to\infty$ so this is impossible.

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Infinitely many. Consider $3$ - term $AP$. There is assertion that there are infinitely many twin primes. The last or the first term will always be divisible by $3$.

Example:

$29, 31, 33$

$27, 29, 31$

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