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

Colin
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    The numerators seem to be given by A005867. I couldn't find a matching sequence for the denominators. – EuYu Nov 15 '12 at 11:57
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    The denominators above $105$ are half the primorials it seems. – EuYu Nov 15 '12 at 12:02
  • You can't just fix the fourth denominator, every subsequent denominator has to be doubled. – EuYu Nov 15 '12 at 12:28
  • 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? – EuYu Nov 15 '12 at 12:33
  • Yes, it is 2, 23, 235, 2357, 235711, 23571113, 2357111317, 235711131719. Sorry I don't know how to change the format so it's easier to read. – Colin Nov 15 '12 at 12:38
  • 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)) – Colin Nov 15 '12 at 12:43
  • Sorry I had miss coppyed it – Colin Nov 15 '12 at 12:49
  • 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. – DonAntonio Nov 15 '12 at 13:23

2 Answers2

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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.

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

EuYu
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