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I have a series of numbers (starting from index n=1): $$\frac{1}{8},\frac{1}{192},\frac{1}{9216},... $$

I notice that all of the denominators are divisible by 8, but I can't seem to find a general rule for this series. Can someone point me in the right direction?

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    $\begingroup$ The Online Encyclopedia of Integer Sequences. $\endgroup$ – Gerry Myerson Dec 8 '13 at 5:32
  • $\begingroup$ After dividing by $8$, you have $\frac 11,\frac 1{24},\frac 1{1048},\dots$. Do you have more terms? $\endgroup$ – abiessu Dec 8 '13 at 5:32
  • $\begingroup$ @abiessu, check your arithmetic. $\endgroup$ – Gerry Myerson Dec 8 '13 at 5:32
  • $\begingroup$ @GerryMyerson : It is not showing anything for $8,192,9216$.. It is showing some results for $8,192$ but the next term is $92160$.. I am afraid there is a typo in the question... not very sure though.. $\endgroup$ – user87543 Dec 8 '13 at 5:33
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    $\begingroup$ Rule of thumb: if the sequence is in the Online Encyclopedia of Integer Sequences, then you can find it there yourself. If it's not in the OEIS, and we are asked, without any context, how it was constructed, then most of us will be hard-pressed to care. It's nothing personal; it's just that this sort of question tends to be for IQ tests and other such instruments, not the actual practice of mathematics. $\endgroup$ – dfeuer Dec 8 '13 at 5:47
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HINT:

$8=2^3,192=2^6\cdot3,9216=2^{10}\cdot3^2$

Now the $n$ th term of the powers of $3$ will be the $n$th term of $0,1,2$ which is $n-1$

the $n$ th term of the powers of $2$ will be the $n$th term of $\displaystyle3=1+2,6=1+2+3,10=1+2+3+4$ which is $\displaystyle1+2+\cdots+n+n+1=\frac{(n+2)(n+1)}2$

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    $\begingroup$ "will be" should be "might be". Frankly there is nothing you can say for certain with so little data. $\endgroup$ – Marc van Leeuwen Dec 8 '13 at 5:43
  • $\begingroup$ @MarcvanLeeuwen, when (with how much data) can we really say for certain? $\endgroup$ – lab bhattacharjee Dec 8 '13 at 5:50
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    $\begingroup$ Never, of course. This kind of question is just not suitable for this site (but it's fine for OEIS). $\endgroup$ – Marc van Leeuwen Dec 8 '13 at 5:55
  • $\begingroup$ That is why I voted to close as "primarily opinion-based". In my opinion, these are the values at $1$, $2$, and $3$ of a certain quadratic equation. $\endgroup$ – dfeuer Dec 8 '13 at 5:56
  • $\begingroup$ Well, also $$8\;,\;192=8\cdot 24\;,\;9216=192\cdot 48\;,\;\;...9216\cdot 72??$$ $\endgroup$ – DonAntonio Dec 8 '13 at 6:41
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Start by factoring the denominators completely: $8=2^3$, $192=2^6\cdot3$, and $9216=2^{10}\cdot3^2$. The powers of $3$ seem to be increasing in a very simple way. Do you recognize $3,6,10$ as part of a familiar sequence of numbers? HINT: Imagine it with a first term of $1$, so that it’s $1,3,6,10$.

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  • $\begingroup$ Well, the three is being raised to a power of $n-1$. I'm not sure what to do with the power of 2 though? $\endgroup$ – Bob Shannon Dec 8 '13 at 5:40
  • $\begingroup$ @Bob: $1=1$, $1+2=3$, $1+2+3=6$, $1+2+3+4=10$, ... $\endgroup$ – Brian M. Scott Dec 8 '13 at 5:42
  • $\begingroup$ I see. How do you take that in to a recurrence formula? It looks like a a factorial except with addition instead of multiplication. $\endgroup$ – Bob Shannon Dec 8 '13 at 5:44
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    $\begingroup$ @Bob: If your terms are numbered starting with $0$, and $a_k$ is the exponent on $2$ in term $k$, then $a_0=3$, and $a_{k+1}=a_k+k+3$; try it and see. $\endgroup$ – Brian M. Scott Dec 8 '13 at 5:47

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