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I was wondering if anyone has come across this sequence and if so if they have a formula for it. $\frac{1}{2},\ \frac{1}{6},\ \frac{2}{30},\ \frac{8}{210},\ \frac{48}{2310},\ \frac{480}{30030},\ \frac{5760}{510510},\ \frac{92160}{9699690},\ \cdots$

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    If $\,d_n\,$ is the n-th denominator, $\,n\geq 2\,$, then $\,d_n=p_nd_{n-1}\,$ , with $\,p_n\,$ the n-th prime.2012-11-15

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This sequence arises in the context of a search for the proportion of nonprimes on intervals of length $p\#. $ The sequence is consistent with the terms of the series:

$S = \frac{1}{2} + \frac{1}{2\cdot 3} + \frac{ 2}{2\cdot3\cdot 5} + \frac{2\cdot 4 }{2\cdot 3\cdot 5\cdot 7} + \frac{}{}...$

The sum of this series is equal to $S = 1-\prod(1 - 1/p_k).$

So the series (and sequence) is very well known but the product lends itself to calculations.

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The sequence, as it stands now, seems to be (starting from $n=1$) $a_n = \frac{\varphi(p_{n-1}\#)}{p_{n}\#}$ where $\varphi$ is Euler's totient function and $p_{n}\#$ is the $n$th primorial (with $p_{0}\#= 1$).