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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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    The numerators seem to be given by [A005867](http://oeis.org/A005867). I couldn't find a matching sequence for the denominators.2012-11-15
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    The denominators above $105$ are half the primorials it seems.2012-11-15
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    You can't just fix the fourth denominator, _every_ subsequent denominator has to be doubled.2012-11-15
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    Yes, those are the primorials. Do you see how all the denominators are doubled after $105$ for that sequence? I _suspect_ that your sequence is $\frac{\phi(p_n\#)}{p_n\#}$ but the numerators and the denominators are shifted. Are you sure this is given correctly?2012-11-15
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    Yes, it is 2, 2*3, 2*3*5, 2*3*5*7, 2*3*5*7*11, 2*3*5*7*11*13, 2*3*5*7*11*13*17, 2*3*5*7*11*13*17*19. Sorry I don't know how to change the format so it's easier to read.2012-11-15
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    I believe the whole series can be written as: where S(n) is the prime number series ((S(n-1) -1) *(S(n-2)-1)*(S(n-3)-1)*...(S(1)-1) ) / (S(n)*S(n-1)*S(n-2)*...S(1))2012-11-15
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    Sorry I had miss coppyed it2012-11-15
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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

2 Answers 2

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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 $p_{0}\#= 1$).