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Let's have the following sequence of natural numbers $4, 5, 7, 9, 11, 14, 16, 20, 22, 27, 29, 35,...a_n,a_n+1$. Does anyone know which two consecutive terms of this sequence have a ratio $(a_{n+1}:a_n)≈(5:1)$?

Informative addendum:

This problem derives from the following poem, written on a marble tablet that was found recently in archeological excavations in the city of Larissa, Greece.

Oh wonderful Goddess of Luck!

Show me how to make a system of lottery tickets where, for every one winner there are five losers.

She answered, "Start by issuing 9 tickets with 5 losers and 4 winners; the tickets of the winners should have the same combination with 6 different numbers. Then, issue 16 tickets with 9 losers and 7 winner; the tickets of the winners should have the same combination with 8 different numbers. Repeat for 14/11, 20/16, 27/22... losers versus winners, with 10, 12, 14 different numbers for the winning combinations, respectively. If you keep doing this continuously, your wish will be fulfilled."

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    I don't find the sequence in [OEIS](http://oeis.org/). Do you have a general rule for $a_n$? What does this have to do with lottery tickets?2012-11-26
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    @Ross I didn't either. It looks like the difference between consecutive numbers: 1 2 2 2 3 2 4 2 5 2 6 2...2012-11-26
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    Wolfram finds a generating function: http://www.wolframalpha.com/input/?i=4%2C5%2C7%2C9%2C11%2C14%2C16%2C20%2C22%2C27%2C29%2C35...2012-11-26
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    Dear Ross Millikan.The challenge is to find how this sequence is formulated. As for the lottery tickets I will put an explanatory addendum.2012-11-26
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    Accepting amWhy's characterization, the largest ratio I find between successive terms is $\frac 75=1.4$. I wouldn't call that close to $5$.2012-11-26
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    -1 Mushing the number of winning, losing, and total tickets in each round makes the sequence much harder to understand. **Please ask the real question the first time.** Now it sounds like you are looking for how many terms are needed to make $5 \approx \frac 54 \cdot \frac 97 \cdot \frac {14}{11} \cdot \frac {20}{16} \ldots$ which is not what you asked at all. I also don't know what to make of the $6, 8, \ldots $different numbers.2012-11-26
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    Dear Ross Millikan.This problem is horrifically difficult.I am going to give the formulas from which the terms of the sequence are obtained.a_n=[a^2-a+2]/2 and a_n+1=[a^2+a-2]/2.Good luck.2012-11-26
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    interestingly, $9 = 3^2$, $16 = 4^2$ ... note that the sums of the numerator and denominator: 14/11: 14 + 11 = 25 = $5^2$, 20/16: 20 + 16 = 36 = $6^2$, 27/22: 27 + 22 = $7^2$...numerators increasing 14 + 6 = 20, 20 + 7 = 27, denominators 11 + 5 = 16, 16+6 = 22. Not that this means anything. I certainly do not understand what your question is! $9 = 3^2$, $16 = 4^2$2012-11-26
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    Dear amWhy.Start from the relation kPn=k!/[k-m]! this formula gives the permutations of k elements m at the time.Keep m=2 aways and k increasing from 3 to infinity.From there is up to you to do the rest of the computations.2012-11-26

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